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PRACTICAL 
DESCRIPTIVE  GEOMETRY 


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PRACTICAL 
DESCRIPTIVE  GEOMETRY 


BY 

WILLIAM  GRISWOLD  SMITH,  M.  E. 

ASSISTANT   PROFESSOR   OF   DESCRIPTIVE   GEOMETRY   AND 
KINEMATICS,    ARMOUR   INSTITUTE    OF    TECHNOLOGY 


SECOND  EDITION 
REVISED,  ENLARGED  AND  RESET 
FIFTH  IMPRESSION 


McGRAW-HILL  BOOK  COMPANY,  INC. 

NEW  YORK:    239   WEST  39TH  STREET 

LONDON:    6  &  8  BOUVERIE  ST.,  E.  C.  4 

1916 


COPYRIGHT,  1912,  1916,  BY  THE 

BOOK  COMPANY,  INC. 


SECOND  EDITION 

Fir  si  printing,  March,  1916 
Second  printing,  August,  1916 
Third  printing,  December,  1917 
Fourth  printing,  November,  1918 
Fifth  printing,  January,  1920 


THE  MAPLE  PRESS  YORK  PA 


Gaelogy 
Library 

QA 

. 

. 

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PREFACE  TO  THE  SECOND  EDITION 

The  first  edition  of  this  book,  published  in  1912,  represented 
the  author's  views  on  the  presentation  of  Descriptive  Geometry, 
so  as  to  arouse  the  student's  interest  in  the  subject  without 
sacrificing  rigorous  training  in  the  principles. 

In  three  years  the  text  has  been  used  in  many  schools,  and  the 
present,  second  edition,  aims  to  embody  the  suggestions  of  many 
leading  teachers  of  the  subject,  who  have,  at  the  request  of 
the  author  and  publishers,  expressed  their  views  as  to  this 
development. 

The  main  idea  of  the  first  edition,  that  of  "practicality,"  has 
been  adhered  to  and  amplified.  The  most  important  changes 
are: 

1.  The  division  into  more  chapters. 

2.  A  more  logical  sequence  for  certain  portions  of  the  text. 

3.  The  notation  of  points  and  planes. 

4.  Graphic  layouts  for  exercises. 

5.  More  exercises  of  a  practical  nature,  embracing  more  lines 
of  engineering  endeavor. 

6.  A  number  of  new  solutions,  utilizing  auxiliary  planes  more 
frequently. 

7.  A  better  system  of  perspective  drawing. 

8.  A  chapter  on  the  intersections  of  plane  solids. 
Although  yielding  to  the  request  of  several  teachers  for  the 

inclusion  of  graphic  layouts  of  exercises,  the  author  still  believes 
that  dimensioned  layouts  are  of  great  value  despite  the  amount 
of  time  expended  on  the  layout  itself.  With  both  of  these,  how- 
ever, in  practically  unlimited  quantity,  the  instructor  may  decide 
the  question  for  himself.  The  layouts  of  both  kinds  have  been 
made  so  flexible,  by  interchanging  and  substituting  data,  that 
the  number  of  variations  is  almost  incalculable.  This  removes 
the  objection,  expressed  by  some  teachers,  to  a  definite  and 
limited  number  of  exercises,  the  solutions  of  which  are  sometimes 
preserved  by  student  organizations.  No  such  collection  of  solu- 
tions will  benefit  future  classes,  as  the  exercises  can  be  made 
entirely  different  each  succeeding  year. 


VI  PREFACE 

As  to  the  treatment  of  Perspective  (how  and  how  much),  the 
author  received  a  large  amount  of  constructive  though  conflicting 
advice,  mainly  from  teachers  of  architecture.  The  writer's  choice 
is  "Conical  Projection,"  a  process  that  is  sane,  in  that  it  is  a 
direct  application  of  the  principles  of  Descriptive  Geometry,  and 
is  easy  and  swift  in  operation. 

It  has  been  previously  stated  that  much  interest  was  mani- 
fested in  the  revision  of  this  work  by  prominent  teachers  in  all 
parts  of  the  country,  and  their  combined  opinions  moulded  it 
into  its  present  form.  The  writer  wishes  to  convey  his  thanks 
to  all  who  assisted  in  this  way.  For  special  criticisms  and  mate- 
rial he  acknowledges  indebtedness  to  Professor  Thomas  E.  French, 
Ohio  State  University;  Professor  Howard  Moore,  Colorado 
College;  Professor  Frederick  C.  Biggin,  Alabama  Polytech- 
nic Institute;  Professor  Robert  H.  McNeilly,  Vanderbilt 
University;  Professor  Chace  Newman,  Michigan  Agricultural 
College;  Professor  H.  B.  Monges,  University  of  California;  Mr. 
Charles  R.  Swineford,  Armour  Institute  of  Technology. 

W.  G.  S. 

ARMOUR  INSTITUTE  OF  TECHNOLOGY, 
Chicago,  April,  1916. 


PREFACE  TO  FIRST  EDITION 

In  presenting  a  new  text-book  on  Descriptive  Geometry,  the 
writer  is  fully  aware  of  the  excellence  of  existing  treatises,  and 
appreciates  what  they  have  contributed  toward  a  higher  standard 
of  technical  training.  The  authors  of  these  treatises,  pursuing 
their  several  courses,  have  carried  the  development  of  the  subject 
to  a  high  plane. 

However,  in  spite  of  the  indisputable  excellence  of  many  of  the 
text-books,  they  seem  to  have  failed  to  arouse  the  interest  of 
the  student,  partly  by  ignoring  the  practical  applications,  and 
partly  by  making  only  a  slight  attempt  to  present  the  subject 
attractively.  Some  of  the  books  are  valuable  only  for  reference 
and  are  useless  in  the  class  room;  others  are  incomplete  in  the 
essentials;  some  are  faddish,  emphasizing  certain  features  and 
treating  the  rest  inadequately,  while  even  the  best  convey  to  the 
student  only  a  very  slight  idea  of  the  practical  value  of  the  sub- 
ject. This  has  given  rise  to  the  belief,  prevailing  almost  uni- 
versally in  the  student  body,  that  Descriptive  Geometry  is  merely 
a  disciplinary  study,  having  little  or  no  relation  to  the  life  work 
of  the  individual. 

The  aim  of  the  writer  has  been,  therefore,  to  present  the 
subject  to  the  student  in  a  simple  manner,  as  progressively  as 
possible,  reminding  him  constantly  of  the  relation  which  exists 
between  Descriptive  Geometry  and  Practical  Drafting;  and  to 
avoid  needless  difficulty  by  using  language  and  directions  of  the 
greatest  possible  clarity.  The  writer  believes  that  a  thorough 
knowledge  of  the  subject  is  achieved  not  through  much  study  of 
the  text,  but  by  working  exercises.  To  this  end  he  has  provided 
a  large  number  of  exercises,  scattered  through  the  text,  of  con- 
siderable variety  and  capable  of  infinite  multiplication  by  the 
clever  instructor. 

An  examination  of  the  subject  will  reveal  the  following  features 
and  innovations: 

1.  A  thorough  drill  in  fundamentals. 

2.  Repetitions  of  statements  for  the  sake  of  emphasis. 

3.  Notation  comprehensive,  yet  reduced  to  its  lowest  terms. 

4.  Analyses  separated  from  proofs. 

5.  Tabulated  order  of  the  operations  in  each  analysis  and 
construction. 


viii  PREFACE 

6.  Exercises  to  the  number  of  860,  of  which  about  one-fourth 
are  such  as  may  be  met  in  actual  practice.     Most  of  these  are 
dimensioned  for  a  space  of  standard  size. 

7.  Notes  on  exceptions,   checks  on  solutions,  special  cases, 
hints,  and  rules  presented  wherever  they  seem  necessary  or 
advisable. 

8.  Brief  treatment  only  of  such  special  subjects  as  Shades  and 
Shadows,  Perspective,  etc. 

9.  The  original  seventeen  "Point,  Line,  and  Plane"  problems 
expanded  to  forty-three. 

10.  New  treatment  of  surfaces,  less  importance  being  attached 
to  passing  tangent  planes,  and  more  importance  to  plane  sections, 
intersections,  developments,  and  practical  applications.     For  ex- 
ample, there  are  about  thirty  exercises  which  are  based  on  prac- 
tical illustrations  of  the  various  warped  surfaces,  showing  actual 
uses  for  all  the  listed  varieties. 

The  expansion  of  the  problems  noted  in  feature  (9)  is  justified, 
the  writer  believes,  in  order  to  fully  cover  the  ground,  and  to 
provide  the  necessary  steps  in  the  development  of  the  subject. 
Some  of  them  are  special  cases  and  some  require  little  or  no 
change  in  the  analysis  from  those  of  preceding  problems. 

Students  undertaking  the  study  of  Descriptive  Geometry 
should  have  a  fair  knowledge  of  Mechanical  Drawing,  at  least  as 
much  as  is  usually  included  in  High  School  courses.  The  mini- 
mum prerequisite  should  amount  to  the  work  contained  in  the 
first  eight  chapters  of  French's  "Engineering  Drawing,"  or  in 
Reid's  "Mechanical  Drawing"  entire. 

In  preparing  this  text-book  the  writer  has  consulted  most  of 
the  standard  works  on  the  subject,  and  desires  to  make  special 
acknowledgment  for  ideas  and  information  to  the  excellent 
books  by  MacCord,  Church  and  Bartlett,  Phillips  and  Millar, 
and  Randall.  Most  of  the  exercises  are  original,  but  many  have 
been  adapted  from  such  admirable  books  of  exercises  as  those 
by  Professors  Hood,  Marshall,  and  Fishleigh,  as  well  as  the  prob- 
lem files  of  Armour  Institute  of  Technology,  and  the  books  of 
the  American  School  of  Correspondence. 

The  writer  wishes  to  thank  his  colleagues,  Robert  V.  Perry, 
Henry  L.  Nachman,  John  S.  Reid,  and  Charles  R.  Swineford,  of 
the  Faculty  of  Armour  Institute  of  Technology,  for  valuable 
suggestions  and  encouragement. 

WILLIAM  GBISWOLD  SMITH. 
ARMOUR  INSTITUTE  OF  TECHNOLOGY, 
Chicago,  June  20,  1912. 


CONTENTS 

PAGE 

PREFACE v 

CHAPTER  I 

FUNDAMENTAL  PRINCIPLES 1-23 

Definitions — Classification  of  Projections — The  Planes  of 
Projection — The  Four  Dihedral  Angles — Theorems — Desig- 
nation and  Notation  of  Points — Drawing  Board  Layout — 
Profile  Projections — Representation  of  Planes — Theorems 
Relating  to  Lines  and  Planes — Exercises. 

CHAPTER  II 

PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES 24-125 

Auxiliary  Planes  of  Projection — Locating  Points  and  Lines 
in  Planes — Revolution  of  Points  about  Axes — True  Length 
of  Lines — Measurement  of  Angles — Figures  in  Oblique 
Planes — Intersections  of  Planes — Projections  of  Oblique 
Solids — Parallel  Plane  Problems — Measurement  of  Dihedral 
Angles — Piercing  Points  of  Lines  and  Planes — Perpendicu- 
lars to  Planes — Special  Problems;  The  Common  Perpen- 
dicular, The  Guide  Pulley,  Reflections — Practical  Exercises. 

CHAPTER  III 

INTERSECTIONS  OP  PLANE  SOLIDS 126-139 

Plane  Sections  of  Pyramids  and  Prisms — Developments — 
Exercises — Intersections  of  Plane  Solids — Exercises. 

CHAPTER  IV 

SHADES  AND  SHADOWS 140-148 

Definitions — Method  of  Determining  Shadows  of  Points, 
Lines,  Plane  Figures,  and  Solids — Exercises. 

CHAPTER  V 

CURVED  LINES 149-168 

Definitions — Projections  of  Curves — Tangents  and  Normals 
— Rectification  of  Curves — Conic  Sections — Gearing  Curves 
— Space  Curves — Problems  relating  to  the  Ellipse,  Hyper- 
bola, Parabola,  Cycloid,  Involute,  Spiral  of  Archimedes 
and  Helix — Exercises. 

ix 


X  CONTENTS 

CHAPTER  VI 

PAO« 

SINGLE  CURVED  SURFACES 169-202 

Definitions  and  Classification — Development — The  Cone — 
Conical  Elbows — Exercises — The  Cylinder — Pipe  Elbows — 
Exercises — The  Convolute — Exercises. 

CHAPTER  VII 

WARPED  SURFACES 203-218 

Methods  of  Generation — The  Helicoid — The  Hyperboloid  of 
Revolution  of  One  Nappe — The  Hyperbolic  Paraboloid — 
Minor  Warped  Surfaces,  The  Cylindroid,  Conoid,  Cow's 
Horn,  Warped  Cone — Development  of  Warped  Surfaces — 
Exercises. 

CHAPTER  VIII 

DOUBLE  CURVED  SURFACES 219-228 

Double  Curved  Surfaces — Surfaces  of  Revolution — Merid- 
ian Sections — Right  Sections — Representation — Problems 
relating  to  the  Sphere,  Torus,  Ellipsoids,  etc.  Plane  Sec- 
tions— Developments — Exercises. 

CHAPTER  IX 

INTERSECTIONS  AND  DEVELOPMENTS  OF  ALL  SURFACES     ....  229-242 
Rule — Simplest  Lines  Cut  by  Planes — Problems  in  Intersec- 
tions and  Developments  of  Various  Surfaces — Concentric 
Spheres  Method — Exercises. 

CHAPTER  X 

PICTORIAL  PROJECTION 243-256 

Perspective — Conical  Projection — Construction  of  Perspec- 
tive from  a  Working  Drawing — Perspective  of  a  Bungalow 
— Exercises — Pseudo  Perspectives;  Isometric,  Cavalier,  and 
Cabinet  Projection — Isometric  Drawing — Exercises. 


PRACTICAL  DESCRIPTIVE 
GEOMETRY 


CHAPTER  I 

DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS, 
PRELIMINARY  PRINCIPLES  AND  EXERCISES 

1.  Definition.  —  Descriptive  Geometry  is  the  Science  of 
Drafting. 

Drafting  is  the  Art  of  representing  objects  on  paper,  or  other 
flat  surface,  by  means  of  projections  of  the  object  on  imaginary 
planes. 


FIG.  1. — The  projection  of  a  point.      FIG.  2. — Orthographic  projection. 


FIG.  3. — Oblique  projection. 

Projections  are  representations  of  objects  made  by  imaginary 
lines,  called  projectors,  running  (or  projected)  from  the  important 
points  of  the  object  to  the  plane  on  which  the  image  is  to  be  made. 

1 


2  PRACTICAL  DESCRIPTIVE  GEOMETRY 

CLASSIFICATION  OF  PROJECTIONS 

Orthographic  Projection:  representation  of  objects  projected 
perpendicularly. 

Perspective  or  Scenographic  Projection:  representation  of 
objects  projected  by  converging  lines. 

Oblique  Projection:  representation  of  objects  projected  by 
parallel  oblique  lines. 

Single  Plane  Projections:  representation  of  objects  in  various 
ways  for  pictorial  purposes.  Several  styles  are  in  use  and  are 
treated  in  Chap.  X.  They  are  mechanical  variants  on  perspec- 
tive and  are  effected  through  various  devices  of  orthographic  and 
oblique  projection,  and  by  artificial  foreshortening. 

2.  Discussion. — Orthographic  Projection  is  almost  universally 
used  in  "working  drawings"  and  "machine  sketches;"  hence 


FIG.  4. — Perspective. 

it  is  the  most  important  class  of  projection.  To  thoroughly 
understand  its  significance,  let  us  consider  the  difference  between 
a  picture  and  a  working  drawing. 

A  picture,  or  perspective,  is  a  representation  of  an  object, 
group,  or  scene,  as  it  appears  to  the  eye.  Rays  of  light  from  the 
object  reach  the  eye  in  converging  lines.  See  Fig.  4.  A  is  the 
eye  of  the  observer;  B  and  C  are  objects  of  equal  size,  but  at 
different  distances  from  A;  D  is  a  plane  of  projection  between  the 
objects  and  the  eye.  Converging  rays  (shown  by  the  dotted 
lines)  project  the  objects  on  D  (piercing  it  at  certain  points). 
These  points  determine  the  size  of  the  pictures  of  B  and  C,  and 
the  images  B'  and  C'  are  the  pictures.  As  B  is  the  nearer,  B'  is 
larger  than  C'.  Thus  distant  objects  appear  smaller  than  nearer 
ones  of  equal  and  often  smaller  size.  By  means  of  this  fact  and 
the  knowledge  of  the  observer  of  the  comparative  sizes  of  the 
objects,  a  picture  shows  the  entire  composition  of  a  scene. 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS       3 

The  working  drawing  takes  no  such  account  of  distances.  See 
Fig.  5.  Here  the  rays  are  all  parallel,  and  the  projections  are  of 
equal  size  to  the  objects,  be  the  latter  near  or  remote.  In  ortho- 
graphic projection,  therefore,  the  eye  must  be  considered  to  be 
at  an  infinite  distance.  If  the  effect  of  a  finite  viewpoint  should 
be  desired,  the  eye  would  have  to  be  shifted  (as  A,  Fig.  5)  to  the 
line  of  each  ray.  This,  of  course,  is  an  artificial  condition,  and 
shows  that  a  working  drawing  is  not,  and  cannot  be,  a  correct 
visual  image  of  the  object. 

This  fact  would  seem  to  be  a  disadvantage  against  the  use  of 
orthographic  projection,  but  for  manufacturing  and  construction 
work,  the  orthographic  is  much  the  more  useful  and  universal 


FIG.  5. 

in  its  adoption  than  perspective,  and  much  easier  for  the  drafts- 
man to  use. 

On  account  of  the  complexity  of  many  machines  and  structures, 
it  is  necessary  to  show  many  dimensions  in  many  directions,  inte- 
rior dimensions,  thicknesses,  and  irregularities,  which  are  all  but 
impossible  in  a  picture.  This  is  accomplished  by  projecting  the 
object  orthographically  on  two  or  more  imaginary  planes,  called 
the  Planes  of  Projection. 

THE  PLANES  OF  PROJECTION 

3.  The  two  principal  planes  are  the  Vertical  and  Horizontal. 

The  Vertical  Plane  (which  we  shall  hereafter  write  V)  is  a  plane 
normal  to  the  earth's  surface;  that  is,  perpendicular  to  the 
apparent  surface  of  the  earth.  The  Horizontal  Plane  (hereafter 
written  H)  is  a  plane  perpendicular  to  any  vertical  line;  that  is, 
it  is  parallel  to  the  apparent  surface  of  the  earth.  H  and  V  are 
perpendicular  to  each  other.  The  projections  on  these  two  planes 
determine  the  relative  position  of  any  two  or  more  points  in  space. 


4  PRACTICAL  DESCRIPTIVE  GEOMETRY 

There  are  conditions  arising  often,  which  render  a  third  plane 
desirable,  and  indeed  necessary,  in  the  solution  of  certain  prob- 
lems. This  third  plane  is  perpendicular  to  both  H  and  F,  and  is 
called  the  Profile  Plane,  which  will  always  be  written  P. 

Any  number  of  jauxiliary  planes,  which  may  be  parallel,  per- 
pendicular, or  oblique  to  H  or  V,  may  be  used,  according  to  the 


FIG.  6. — Machine  part  projected  orthographically. 

complexities  of  the  problem.  (In  the  architect's  plans  of  a  big 
building,  two  different  vertical  projections  may  be  needed,  two 
profiles,  a  horizontal  projection  for  every  floor,  basement,  and 
roof,  auxiliary  projections  of  certain  oblique  features,  sectional 
interior  projections,  and  a  hundred  "detail"  projections  of  doors, 
windows,  staircases,  elevators,  etc.,  all  in  orthographic  projection. 
Besides  these,  it  is 'customary  to  make  a  perspective  of  the  pro- 
jected building  with  colors,  trees,  grass,  walks,  etc.  The  average 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS       5 

purchaser  cannot  comprehend  working  drawings,  while  the  pic- 
ture is  of  little  use  to  the  workman.  Perspectives  are  for  the 
customers,  orthographic  projections  for  the  workmen.) 


FIG.  7. — Projections  of  machine  part  in  the  flat. 

4.  It  is  obvious  that  it  would  be  awkward  and  impracticable  to 
have  a  drawing  board  made  with  horizontal,  vertical,  and  profile 
sides,  and  the  respective  projections  made  on  them.  Not  only 
would  such  an  arrangement  be  unwieldy,  but  it  is  unnecessary. 
Fig.  8  shows  two  planes,  H  and  V,  intersecting,  as  they  must,  in 
a  line,  called  the  Ground  Line  (designated  hereafter  GL). 

By  revolving  V  about  GL  as  an  axis  or  hinge,  it  will  be  put  into 
coincidence  with  H,  and  the  H-  and  F-projections  can  be  drawn 
on  one  surface.  Similarly,  P  can  be  rotated  into  the  same  surface 
about  its  intersection  either  with  H  or  V.  For  mere  solution  of 
problems,  it  is  immaterial  whether  the  P-projection  is  revolved 
into  H  or  V,  but  in  practical  work  it  is  customary  to  revolve  it 
into  V;  therefore  in  this  work  the  profile  will  always  be  rotated 
about  its  F-trace  into  V. 

THE  FOUR  DIHEDRAL  ANGLES 

6.  It  will  be  seen  from  Fig.  8  that  the  planes  of  projection  do 
not  end  at  GL.  In  fact,  they  are  unlimited  in  extent,  and  divide 


6 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


space  into  four  dihedral  angles,  with  GL  as  their  common  inter- 
section. In  folding  into  the  fiat,  therefore,  no  portion  of  the 
drawing  board  is  monopolized  by  either  H  or  V.  There  is  a  por- 
tion of  H  above  GL,  and  a  portion  below  GL.  There  is  a  portion 
of  V  above  GL,  and  a  portion  below  GL.  (This  circumstance  will 
be  found  essential  to  the  solution  of  certain  problems  of  abstract 
character.  In  practical  work  the  different  projections,  or  views, 


FIG.  S. 

as  they  are  often  called,  are  entirely  separated  from  each  other, 
and  definite  spaces  are  reserved  for  H ,  V,  and  P.) 

The  four  dihedral  angles,  in  any  of  which  objects  may  be  imagined,  are 
numbered,  as  shown  by  the  Roman  numerals  in  Fig.  8. 
First  Angle,  in  front  of  V,  above  H; 
Second  Angle,  behind  V,  above  H; 
Third  Angle,  behind  V,  below  H; 
Fourth  Angle,  in  front  of  V,  below  H. 

The  first  angle  (designated  in  this  work  as  7)  is  largely  used  in 
our  abstract  problems,  on  account  of  its  accessibility  and  the  ease 
of  presentation  by  the  teacher  to  the  class,  when  giving  a  solution 
of  a  problem  in  space. 

The  third  angle  (III}  is  commonly  used  in  practical  work, 
for  the  reason  that  the  H-p rejection  is  usually  made  looking 
down  on  the  object,  and  besides  for  this  angle  the  profile  comes 
out  right-handed  and  detached,  whereas  the  usual  method  of 
swinging  a  profile  from  a  first-angle-projection  will  either  bring 
the  object  out  left-handed,  or  mixed  up  with  the  other  views. 
Many  of  our  problems  are  given  in  777. 

6.  Commercial   Terminology   of   Projections. 

As  the  terms  F-projection,  H-p  rejection,  etc.,  are  not  descrip- 
tive to  non-professionals,  certain  terms  are  used  by  draftsmen. 

The  F-projection  is  often  called  the  Elevation,  Front  View, 
Rear  View,  or  Sectional  Elevation,  as  the  case  may  be. 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS       7 

The  //-projection  is  called  the  Plan,  Top  View,  Bottom  View, 
or  Sectional  Plan,  as  the  case  may  be. 

The  P-projection  is  called  the  Profile,  End  View,  Right  or 
Left  End  View,  or  Profile  Section,  as  the  case  may  be. 

THE  REPRESENTATION  OF  POINTS,  LINES,  AND  PLANES 

7.  Points. — It  is  obvious  that  the  projection  of  a  point  on  one 
plane  does  not  determine  that  point.     For  example,  let  a  point 
on  the  floor  be  the  projection  of  some  point  in  the  room.     How 
high  is  that  point?     It  may  be  any  distance.     The  point  may  be 
the  projection  itself,  or  the  point  may  be  on  the  ceiling  above,  or 
anywhere  in  between.     To  locate  the  point  in  space  by  one  pro- 
jection, it  is  necessary  to  add  its  distance,  positive  or  negative, 
from  the  projection.     A  better  way  (the  draftsman's  way)  is  to 
locate  the  point  graphically.     By  projecting  that  point  on  the 
wall,  its  height  is  determined  and  shown. 

This  definitely  locates  a  point,  for,  if  perpendicular  lines  be 
erected  from  each  projection  of  a  point,  these  perpendiculars 
will  pass  through  the  point.  As  there  can  be  but  one  intersection 
of  the  perpendiculars,  the  point  is  absolutely  located  in  space  by 
its  projection  on  two  planes.  The  perpendicular  lines  from  the 
point  to  the  planes  of  projection  are  called  projectors. 

THE  RELATION  OF  PLANE  AND  SOLID  GEOMETRY  TO  DESCRIPTIVE 

GEOMETRY 

8.  All  theorems  in  Plane  and  Solid  Geometry  are  applied  to 
Descriptive  Geometry  to  a  greater  or  less  extent.     There  are,  how- 
ever, a  number  of  theorems  of  outstanding  importance,  to  which 
reference  must  be  made  with  considerable  frequency.     A  list  of 
these  is  herewith  submitted.     The  student  need  not  be  required 
to  prove  them,  but  should  be  able  to  give  the  conclusion  to  every 
hypothesis,  and  to  quote  his  references  on  occasion. 

PLANE  AND  SOLID  GEOMETRY  THEOREMS 

1.  The  opposite  sides  of  a  parallelogram  are  equal. 

2.  Triangles  are  equal: 

(1)  If  three  sides  of  one  are  equal  to  three  sides  of  the  other, 
each  to  each. 


8  PRACTICAL  DESCRIPTIVE  GEOMETRY 

(2)  If  two  sides  and  the  included  angle  of  one  are  equal  to 
two  sides  and  the  included  angle  of  the  other. 

(3)  If  one  side  and  the  adjacent  angles  of  one  are  equal  to 
one  side  and  the  adjacent  angles  of  the  other. 

3.  Right  triangles  are  equal  if  any  side  and  an  acute  angle  of 
one  are  equal  to  the  corresponding  side  and  angle  of  the  other. 

4.  If  the  two  points  of  a  line  lie  in  a  plane,  the  line  lies  entirely 
in  the  plane. 

5.  If  a  line  is  perpendicular  to  a  plane,  it  is  perpendicular  to 
all  the  lines  in  the  plane  that  intersect  it. 

6.  Two  straight  lines  perpendicular  to  the  same  plane  are  par- 
allel to  each  other. 

7.  If  two  parallel  planes  are  cut  by  a  third  plane,  the  inter- 
sections are  parallel. 

8.  If  a  line  external  to  a  plane  is  parallel  to  any  line  in  that  plane 
it  is  parallel  to  the  plane. 

9.  If  a  line  is  parallel  to  two  intersecting  planes,  it  is  parallel 
to  their  intersection. 

10.  A  line  perpendicular  to  one  of  two  perpendicular  planes 
is  parallel  to  the  other. 

11.  A  plane  perpendicular  to  two  intersecting  planes  is  per- 
pendicular to  their  intersection. 

12.  The  dihedral  angle  between  two  planes  is  measured  by  the 
angle  of  the  lines  cut  from  them  by  a  plane  perpendicular  to 
their  intersection. 

13.  If  a  line  is  perpendicular  to  a  plane,  every  plane  containing 
the  line  is  perpendicular  to  the  plane. 

14.  If  two  planes  are  perpendicular,  a  line  in  one,  perpendicu- 
lar to  their  intersection,  is  perpendicular  to  the  other. 

15.  The  angle  that  a  line  makes  with  a  plane  is  the  angle 
that  it  makes  with  its  projection  on  that  plane. 

16.  Two  lines  parallel  in  space  are  projected  on  any  plane  in 
parallel  lines. 

THEOREMS  RELATING  TO  POINTS 


9.  Theorem  I. — The  H-  and  V-projections  of  a  point  must  lie 
on  a  common  perpendicular  to  GL. 

Proof. — The  H-  and  7-projectors  of  the  point  determine  a 
plane  that  is  perpendicular  to  both  planes,  and  therefore  to  GL. 
(Why?)  The  intersections  of  this  plane  with  H  and  V  will  then 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS       9 

be  perpendicular  to  GL  and  at  the  same  point,  which  proves  the 
theorem. 

In  Fig.  9,  the  point  A,  in  space,  is  projected  on  H  and  V,  in 
aH  and  av  respectively,  so  that  the  projectors  and  their  projections 
avx  and  aHx  form  a  rectangle.  This  fact  also  proves  the  following 
theorem : 


FIG.  9. 

Theorem  II. — The  distance  that  any  point  is  from  H  is  equal 
to  the  distance  from  its  V-projection  to  GL;  the  distance  that 
any  point  is  from  V  is  equal  to  the  distance  from  its  H-projection 
to  GL. 

PROJECTIONS  OF  POINTS  IN  VARIOUS  ANGLES 

By  folding  H  and  V  from  their  natural  positions  "into  the  flat," 
we  find  that  the  projections  of  points  in  the  different  space  angles 
are  differently  located  relatively  to  GL.  The  first  and  third 
angles  are  opened  out,  so  that  the  projections  of  their  points  lie 
on  opposite  sides  of  GL,  whereas  the  second  and  fourth  angles  are 
closed,  locating  their  projections  on  one  side  of  GL. 

A  study  of  Fig.  10  will  show  that  the  first  and  second  angles 
are  above  H  and  on  opposite  sides  of  V,  and  that  the  third  and 
fourth  angles  are  below  H  and  on  opposite  sides  of  V.  The 
first  and  fourth  angles  are  in  front  of  V  and  the  second  and  third 
are  behind  it.  In  this  figure  a  point  is  shown  in  each  of  the  four 
angles  with  arrows  pointing  the  rotation  of  H  into  the  flat. 

Fig.  10a  shows  the  representation  of  these  points  on  the  flat, 
and  the  following  theorem  is  the  direct  deduction  from  the 
arrangement  of  the  various  projections  after  their  revolution. 

Theorem  III. — Points  located  in  front  of  V  have  their  H-pro- 
jections  below  GL.  Points  behind  V  have  their  H-projections 


10 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


above  GL.    Points  above  H  have  their  V-projections  above  GL. 
Points  below  H  have  then*  V-projections  below  GL. 

Theorem  IV. — Points  in  H  have  their  V-projections  hi  GL,  and 
points  hi  V  have  their  H-projections  hi  GL. 


Fourth  Angle 


Second   Angle 


Third  Angle 


FIG.   10. — The  four  dihedral  angles. 


' 


•* 


m 


w  dH 


FIG.  10a. 

Proof. — A  point  in  H  is  zero  distance  from  H,  hence  its  V- 
projection  is  zero  distance  from  GL.     (Theorem  II.) 

DESIGNATION  AND  NOTATION  OF  POINTS 

10.  The  actual  point  in  space  is  designated  by  a  capital,  A,  B, 
C,  etc. 

The  F-projection  is  designated  av,  bv,  cv,  etc. 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS     11 

The  ^-projection  is  designated  aH,  bH,  CH,  etc. 

The  P-projection  is  designed  ap,  bp,  cp,  etc. 

Revolved  Positions. — If  a  point  is  taken  from  its  original  posi- 
tion, and  placed  in  a  new  one,  it  is  designated  by  a  subscript,  as 
aiv,  or  aiH.  If  there  be  a  second  revolution,  it  becomes  auv,  or 
auH,  etc. 

NOTATION  OF  A  POINT 

There  are  three  items  in  the  notation  of  a  point : 

1.  The  distance  from  the  margin,  measured  along  GL  to  the 
right. 

2.  The  distance  above  or  below  GL  (+  or  — )  of  its  V-pro- 
jection. 

3.  The  distance  above  or  below  GL  (+  or  — )  of  its  H -pro- 
jection. 


dv 
|— x 


c- -^ H 


H 

• 


FIG.  11. 

Example. — Draw  on  the  flat  the  projections  of  A(33^  +  1  —  2). 

1.  Lay  off  a  point  on  GL,  3%  in.  to  the  right  of  the  margin. 

2.  Locate  av  1  in.  above  (  +  1)  this  point  on  GL. 

3.  Locate  aH  2  in.  (  —  2)  below  this  point  on  GL. 

Notes. — 1.  Plus  always  means  above  GL,  Minus  always  means  below 
GL. 

2.  First  locate  the  projections  of  the  point  on  paper,  and  then  try  to 
imagine  it  in  space. 

3.  The  unit  for  the  drawing  board  is  1  in.,  and  for  blackboard  work  5  in. 
is  a  good  scale  for  the  exercises  in  this  book. 


LAYOUT  FOR  DRAWING  BOARD 

11.  A  convenient  layout  is  shown  in  Fig.  12.  A  sheet  15  X  20 
in.  is  given  a  1^-in.  margin  on  the  left  for  binding,  and  a  3^-in. 
margin  on  the  other  three  sides,  leaving  a  working  space  of  14  X 


12 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


18  in.     This  space  is  divided  into  six  rectangles,  each  6  X  7  in., 
with  GL  drawn  through  the  middle  of  each. 

Inking  Plates. — It  is  not  advisable  to  ink  plates,  except  the 
border  lines,  ground  lines,  and  lettering.  Good  pencil  work  and 
neat,  legible  lettering  should  be  insisted  on. 


! 

A 

! 

: 

1 

N 

/'" 

„ 

„                , 

>? 

&  —    —>  g 

*o 

V 

i 

1 

t  V 

Y 

—  ICU 

/VAMh  . 

, 

* 

-—  -J 

FIG.  12. 

Projectors. — The  perpendicular  lines  running  from  the  pro- 
jections to  GL  are  called  projectors  and  should  be  light  dotted 
lines.  See  Figs.  10  and  11. 


12. 


EXERCISES  IN  POINT  REPRESENTATION 


Locate  on  the  flat  and  designate  in  space  (see  the  Roman  Numerals,  etc., 
in  Fig.  10a)  the  following  points: 

1.  A(%  -1+2),  BUM  +  M  +  2),  C(2K,   O  +  1),  D(3  +  1  -  1),  E 
(3M  -  2  -  M),  F(4M,  0,  0),  and  G(5M  +  1M,  0). 

2.  K(3A  +  y2  -  2),   L(1H  ~  1  -  2),    M(2M  -  IK,    0),    N(3  -  1  +2), 
0(3H  +2  +  1),  P(4M  +  1,  0),  Q(5K,  0  +  1). 

3.  A(M,  1  behind  F,  2  below  H),  BUM,  in  #,  2  before  F),  C(2H,  1  before 
F,  1  above  #),  D(3,  1  behind  F,  2  above  tf ),  E(3H,  in  (?L),  F(4M,  1 
before  F,  2  above  #),  G(5K,  ia  ^,  2  below  £f). 

4.  K(M,  in  -ff,  2  behind  F),  L(1M,  1  before  F,  2  below  H),  M(2^,  1 
behind  F,  1M  above  #),  N(3,  2  before  F,  1  above  #),  O(3%  in 
P(4M,  1  behind  F,  2  below  H),  Q(5H,  1M  above  //,  in  F). 

Locate  in  the  flat  the  following  points: 

5.  A  (in  IV,  1  to  right,  1  from  F,.2  from  #), 
B(in  /,  2  to  right,  2  from  F,  1  from  #), 
C(in  77,  3  to  right,  1M  from  F,  1M  from  77). 

6.  D(in  77,  1  to  right,  1  behind  F), 

E(in  777,  2  to  right,  2  from  F,  1  from  77), 
F(in  7,  3  to  right,  1  from  F,  1M  from  77). 

7.  G(in  77,  1  to  right,  1M  from  77,  1M  from  F). 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS     13 

K(in  V,  between  ///  and  IV,  2  to  right,  2  from  H), 
L(in  H,  between  7  and  IV,  3  to  right,  1  from  7). 
8.  M(in  IV,  1  to  right,  2  from  H,  2  from  V), 
N(in  III,  2  to  right,  1  from  V,  \YZ  from  H), 
O(in  H,  between  II  and  777,  3  to  right,  2  from  7). 

LINES 

13.  A  straight  line  is  the  path  of  a  point  that  moves  constantly 
in  one  direction.     Any  two  points  therefore  of  a  straight  line  are 
sufficient  to  determine  it.     Also,  the  projection  of  that  line  will 
contain  the  projections  of  the  points.     Until  the  subject  of  curved 
lines  is  taken  up  the  term  "line"  will  be  understood  to  mean  a 
straight  line. 

Theorem  V. — Two  projections  of  a  line  will,  in  general, 
determine  the  position  of  that  line  in  space. 

Proof. — Pass  a  plane  through  the  F-projection  of  some  line, 
and  make  it  perpendicular  to  V.  Also  pass  a  plane  through  the  H- 
projection  of  the  same  line  and  make  the  plane  perpendicular  to 
H.  These  planes  will  intersect  in  a  line,  and  that  line  must  be 
the  given  line. 

Theorem  VI. — If  a  line  is  parallel  to  H,  its  H-projection  is 
parallel  and  equal  to  the  line,  and  its  V-projection  is  parallel 
to  GL. 

Prove  This. — What  would  be  the  statement  if  the  line  were 
parallel  to  V? 

Theorem  VII. — If  a  line  is  parallel  to  H  and  V,  both  of  its 
projections  will  be  parallel  to  GL. 

Theorem  VIII. — If  a  line  is  perpendicular  to  H,  its  H-projec- 
tion will  be  a  point,  and  its  V-projection  will  be  perpendicular 
to  GL. 

What  would  be  the  statement  for  a  line  perpendicular  to  Vf 

Theorem  IX. — If  a  line  is  oblique  to  both  H  and  V,  both  pro- 
jections will  be  oblique  to  GL,  except  when  the  line  lies  in  a 
profile  plane. 

14.  PROJECTIONS  ON  THE  PROFILE  PLANE 

The  projections  of  the  solid  shown  in  Fig.  13  do  not  completely 
identify  the  object,  because  the  ends  (bases)  lie  in  a  profile  plane. 


14 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


The  end  view  (sometimes  called  the  right  section)  might  be  one 
of  many  forms  and  in  different  positions,  as  the  shapes  given  in 

Fig.  14.  Thus  the  profile  is  some- 
times necessary  to  complete  the 
drawing. 

In  drawing  views  in  profile,  the 
right  end  view  is  usually  taken, 
although  sometimes  both  profiles 
are  necessary,  and  some  cases  arise 
where  only  the  left  view  is  needed. 

Taking  the  view  at  the  right,  we  see  the  four  dihedral  angles  as 
shown  in  Fig.  15.  Looking  along  GL  we  see  H  and  V  in  their  true 
relation  (perpendicular)  ,  and  the  four  space  angles  as  indicated  by 
the  Roman  Numerals. 


FIG.  13. 


FIG.  14. 


Assume  P  at  any  convenient  location  on  the  right.  In  the 
regular  views,  mv  gives  the  elevation  and  mH  the  distance  in 
front  of  V.  In  the  profile  view  it  will  be  seen  that  the  same  dis- 
tances obtain.  If  the  point  lies  in  any  other  angle,  the  pro- 


H- 


IV 


*-• 


H  G- 


m 


V 

FIG.  15. 


K>|QO                   y 
•X-Lff >- 

Three  Projections  j 
of  aPoinr. 


FIG.  15a. 


cedure  is  the  same.  The  rotation  of  the  //-projection  will  de- 
termine whether  the  profile  of  the  point  is  to  lie  to  the  right  or 
left  of  P.  If  the  //-projection  is  below  GL,  the  profile  will  be  on 
the  left,  as  in  Fig.  15a,  and  if  it  be  above  GL,  the  profile  will  be 
on  the  right.  The  elevation  is  always  the  same  on  both  V  and  P. 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS     15 


Rule. — Project  from  the  V-projection  horizontally;  project 
from  the  H-projection  to  P,  and  rotate  to  GL,  thence  up  or  down 
until  the  projectors  intersect. 

To  obtain  the  profile  of  a  line  or  plane  figure,  project  the  de- 
fining points  and  join  them  with  lines.  Fig.  16  gives  a  pictorial 
view  of  a  line  and  its  projections.  Fig.  16a  shows  the  same  line 
in  the  flat. 


Ctr- 


FIG.  16. 


Flat  Projection. 
FIG.  16a. 


Theorem  X. — If  a  line  lies  in  V,  its  V-projection  is  the  line 
itself,  and  its  H-projection  is  in  GL. 

Why?     What  would  be  the  statement  for  a  line  in  Hf 

Theorem  XI. — If  a  point,  line,  or  plane  figure  lies  in  P,  its  V- 
projection  is  in  the  V-trace  of  P,  and  its  H-projection  is  in  the 
H-trace  of  P. 

16.  EXERCISES 

-   9.  Draw  the  line  A(l  +  2  -  2^)  B(4  -  2  +  K),  and  tell  where  it  travels. 

10.  Draw  the  line  C(l^  -  2  +  1^)  D(4K  +3-1),  and  tell  where  it 
travels. 

11.  Draw  and  designate  the  lines:  E(l  +  1  -  1)  F(3  +  2  -  3),  G(3^  + 
1  +  1K)K(4^  +2  -y2),L(5  -1  -2)M(5%  -3-1). 

12.  Draw  and  designate  the  lines:  A(l  -  2  +  2)  B(2K  -  M  +  1),  C(3  + 
1  +  1)  D(4  +  2  +  2),  E(4M  -  1  -  3)  F(5^  +1+1). 

Draw  and  designate  the  lines  in  the  following: 

13.  K(l  +  1  -  1)    L(3  +  3  -  1),    M(2^  -2+2)    N(4  -  2  +  1),    and 
0(4  +  2  -  1)    P(5M  +3+3). 

14.  A(l  +  3  +  K)  B(3  +  1  +  y2),  C(2  -  3  -  2)  D(4  -  3  -  1),  E(4^  + 
2-1)  F(4^  +2-3),  and  G(5M  -  1  +  D  K(5^  -1+3). 

15.  K(J^  -1+1)  L(2K  -3  +  1),  M(2  -  H  -  1)  N(4  -  M  -  D,  and 
O(3  +  2  +  1)  P(5  +  2  +  3). 

Tell  what  angles  these  lines  traverse: 

16.  A(l  +  2  -  1)  B(3  -  1  -  3),  C(2  +  3  +  1)  D(4  -  2  -  2),  and  E(4  - 
1  +  2)  F(5K  +2-2). 

17.  G(l  -  1  -  3)    K(2  +  2  +  3),    M(2  -3  +  1)    N(3M  -1-D,    and 
0(4  +  2  -  1)  P(5H  +  1+3). 


16 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


18.  From  A(l  —1+2)  draw  a  line  2  in.  long,  parallel  to  V,  and  inclined 
60°  to  H,  downward.     Does  it  touch  Hf     V?     Draw  one  60°  upward, 
2J^  in.  long.     Where  does  it  terminate? 

19.  From  B(2  -  2  -  1)  draw  a  line  ||  to  H,  inclining  30°  to  V.     How  far  is 
it  on  this  line  from  B  to  V? 

20.  Draw  the  P-projections  of  the  lines  A(l  +  1-2)  B(l  +3-1)  and 
C(2  -  1  +  1)  D(2  -2  +  1).     Locate  the  profile  at  4>£  on  GL. 

21.  Draw  the  P-projections  of  E(l  +  1  +  1)  F(l  -  1  -  2)  and  G(2  +  2  + 
1)  K(2  -  1  +  1).     Profile  4^. 

22.  Draw  the  triangle  L(2  +  1  -  1)   M(2  +  2  +  M)  N(2  -  1  -  1M)  in 
its  true  size.     Profile  4J£. 

23.  What  are  the  sizes  of  the  angles  and  sides  of  the  triangle  K(2  +  1  +  1) 

0(2  -  i  +  iH)  P(2  -  y2  -  2)? 

24.  Draw  the  projections  of  a  prism  parallel  to  H  and  V,  of  2-in.  altitude, 
and  its  base  a  pentagon  inscribed  in  a  1  J^-in.  circle  in  P. 

25.  From  M(l  —  2  —  1)  draw  a  line  ||  to  H,  inclined  45°  to  V,  3  in.  long, 
toward  V  to  the  right.     Also  from  M  draw  a  line  ||  to  P,  30°  to  V,  3  in. 
long.     Where  do  they  terminate? 


REPRESENTATION  OF  PLANES 


L 


FIG.  17. 


(a) 
FIG.  17a. 


16.  A  plane  other  than  H,  V,  or  P  is  shown  by  its  traces,  as  its 
intersections  with  the  three  planes  are  called.  In  most  cases  the 
H-  and  F-traces  are  sufficient,  and  the  P-traee  is  chiefly  used  as 
an  auxiliary  to  assist  in  the  solving  of  problems.  Fig.  17  shows 
an  oblique  plane,  T.  The  plane  T  cuts  V  in  the  line  VT,  called  its 
vertical  trace.  It  cuts  H  in  the  line  HT,  called  its  horizontal 
trace.  Fig.  17a  shows  the  flat  representation  of  the  plane  T. 
VT  is  one  line  of  T  and  HT  is  an  entirely  different  one. 

Theorem  XII. — The  two  traces  of  a  plane  completely  deter- 
mine the  plane. 

Note. — The  student  must  on  no  account  think  that  VT  and  HT  are  pro- 
jections of  the  same  line.  VT  lies  in  V,  hence  its  H-p rejection  is  in  GL. 
Therefore,  if  a  point  is  assumed  in  the  plane  with  its  F-projection  in  VT, 
its  ^/-projection  is  in  GL.  What  is  the  case  then  with  HT? 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS     17 

Theorem  XIII. — The  two  traces  of  any  plane  must  both  be 
parallel  to  GL,  or  they  must  intersect  GL  and  in  one  point. 

Let  the  student  prove  this. 

Note. — Planes  are  unlimited  in  extent,  and  the  traces,  therefore,  do  not 
end  in  GL,  but  continue  indefinitely.  Every  plane,  therefore,  that  is  oblique 
to  GL  occupies  every  space  angle. 

Theorem  XIV. — If  a  plane  is  parallel  to  V,  it  has  but  one  trace, 
the  H-trace,  parallel  to  GL.  What  would  be  true  of  a  plane 
parallel  to  Hf 

Theorem  XV. — The  traces  of  P  are  both  perpendicular  to  GL. 

Theorem  XVI. — A  plane  perpendicular  to  H,  but  oblique  to  V, 
has  its  V-trace  perpendicular  to  GL.  For  example,  see  a  swing- 
ing door  as  it  takes  various  positions.  Such  a  plane  is  called  an 
H-projecting  plane,  for  the  reason  that  all  points  in  it  are  pro- 
jected on  H  in  its  H-trace.  What  would  be  true  of  a  plane  per- 
pendicular to  Vf  See  a  box  lid. 

Theorem  XVII. — If  a  plane  passes  through  GL,  it  can  only  be 
satisfactorily  shown  by  its  P-trace.  Both  H-  and  F-traces  are 
in  GL,  therefore  it  requires  a  point,  or  the  P-trace,  to  determine 
it.  It  is  a  P-projecting  plane.  Why? 

Theorem  XVIII. — Two  parallel  planes  have  parallel  traces  on 
any  plane  of  projection.  Give  a  reason  for  this  from  Solid 
Geometry.  Is  the  converse  always  true?  Take  two  intersecting 
planes  parallel  to  GL. 

Theorem  XIX. — Two  planes  perpendicular  to  each  other  do 
not  in  general  have  their  traces  respectively  perpendicular. 

The  only  exception  to  be  found  is  the  case  of  a  profile  plane  and 
one  parallel  to  GL.  Two  perpendicular  planes  that  are  per- 
pendicular to  H  will  have  their  ^-traces  perpendicular,  but 
their  F-traces  will  be  parallel. 

NOTATION  OF  PLANES 

17.  Planes  are  represented  by  their  traces  and  are  completely 
determined  by  them. 

The  conventional  dot-and-dash  line  is  sufficient  to  identify 
them  in  a  drawing. 

They  are  lettered  VT,  HT,  PT,  VS,  HS,  PS,  etc.,  to  dis- 
tinguish the  V-,  H-,  and  P-traces  of  T,  S,  etc, 
2 


18 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


NOTATION 


When  the  traces  intersect  in  the  problem  limits,  three  points 
are  necessary: 

1.  In  the  F-trace. 

2.  The  GL  intersection. 

3.  In  the  //-trace. 

All  these  points  are  in  H  or  V,  hence 
there  are  zeros  in  the  complete  notation. 
These  zeros  are  understood  and  omitted 
in  our  notation,  as  being  cumbersome. 


(a) 

FIG.    17a. — The    plane 
T  (1  +  2)  5(1  -  1). 


1.  (1+2)  means  a  point  in  VT,  1  in.  to  right,  2  in.  up. 

2.  5  means  the  GL  intersection,  5  in.  to  right. 

3.  (1  —  1)  means  a  point  in  HT,  1  in.  to  right,  1  in.  down. 
Rule. — First. — A  point  in  the  F-trace.    Second. — The  GL  inter- 
section.    Third. — A  point  in  the  H-trace. 

SPECIAL  CASES 

Planes  parallel  to  GL  have  no  GL  intersection  and  their  traces 
are  parallel  to  GL,  and  their  notation  is  shown  in  Fig.  18. 

VS  uX 


i 

/' 

: 

ax^-— 

I 

^s* 

t 

- 

\ 

J 

r                             HS 

^-- 

FIG.  18.— The  plane  S(+  !)»(-  2). 

Fig.  18  shows  the  plane  S  and  its  profile  trace,  which  is  fre 
quently  necessary  with  such  planes. 


FIG.  19.—  The  plane  X(l  +  3)  (5  +  2)(1  - 


-  1). 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS     19 

Planes  parallel  to  H  or  V  have  but  one  trace,  and  that  parallel 
to  GL.  They  require  special  designation,  but  are  seldom  used 
except  as  auxiliary  planes,  and  so  it  is  not  necessary  to  devise  a 
special  notation  for  them. 

OTHER  PLANES 

A  plane  oblique  to  GL,  but  not  crossing  it  within  the  problem 
limits,  must  have  four  points  to  determine  the  two  traces. 

In  this  plane  the  first  two  points  are  in  the  V-trace  and  the 
second  pair  are  in  the  .H-trace.  See  Fig.  19. 

A  plane  through  GL  has  its  V-  and  .//-traces  in  GL,  therefore 
its  P-trace  is  its  best  exposition.  It  can  be  described  also  as 
a  plane  passing  through  GL  making  a  certain  angle  with  the 
horizontal,  or  containing  some  point. 

EXERCISES  IN  REPRESENTATION  OF  PLANES 

18.  Draw  the  traces  of  the  following  planes,  and  state  their  relations 
(parallel,  perpendicular,  oblique)  to  H,  V  and  P. 

Draw  the  traces  with  a  dot-and-dash  line.  In  Ex.  28  and  29,  place  P  at 
4J^,  and  draw  the  P-traces. 

26.  T(3  +  2)  1(1  -  3),  8(5  +  3)  3(3  -  3). 

27.  X(3  +  3)  1(3  +  1),  Y(3M  +  2)  5(4  -  3). 

28.  R(  +  1)  «>  (  +  2),  Q(  -  1)  co  (  +  2).     Draw  the  P-traces. 

29.  S(  -  2)  co  (  -  1),  T(  +  2)  oo  (  -  l).     Draw  the  P-traces. 

30.  X(l  + 1)  (4  +  2)  (1  +  IK)  (4  +  3),  TJ(2  -  3)  4(5K  +  2^). 

31.  Draw  the  traces  of  the  F-projecting  plane  of  the  line  M(l  +2  —  J£) 
N(2>£  +  K  —  1).  and  of  the  ^-projecting  plane  of  the  line  A(3K  - 
K  -  2)  B(5  +  2  +  W). 

32.  Draw  the  traces  of  the  F-projecting  plane  of  C(l  +3  —  1)  D(3  —  1  + 
3)  and  of  the  //-projecting  plane  of  E(3K  +  1-2)F(5  +  1-1). 

33.  Draw  the  traces  of  the  77-projecting  plane  of  G(3  +  3  —  1)  K(l  —  1  + 
1),  and  of  the  F-projecting  plane  of  M(3^  +  1-2)  N(5M  +  3  +  K)- 

34.  Draw  the  P-trace  of  S,  which  contains  GL  and  O(2  +2  —  1),  and  of 
R,  which  contains  GL  and  A (4  +  1  +  1^). 

35.  Draw  the  P-traces  of  two  planes,  T  and  X,  each  containing  GL,  and 
each  inclined  60°  from  the  horizontal,  T  passing  through  I  and  777,  X 
passing  through  77  and  7F. 

Note. — To  find  the  P-trace  of  any  plane,  take  the  plane  P  at  any  conven- 
ient place,  where  it  will  intersect  the  77-  and  F-traces.  Then  treat  the  line 
connecting  these  intersections  (the  P-trace)  as  you  would  any  other  profile 
line,  and  revolve  it  as  in  Art.  14. 

THEOREMS  RELATING  TO  LINES  AND  PLANES 

19.  Theorem  XX. — Two  lines  parallel  in  space  have  parallel 
projections  on  any  plane. 


20 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Theorem  XXI. — Two  lines  perpendicular  in  space  do  not 
ordinarily  have  their  projections  perpendicular. 

Special  Case. — If  one  of  these  lines  is  parallel  to  H,  the  H-pro- 
jections  of  the  two  lines  will  be  perpendicular. 

Fig.  20  shows  a  line  AB,  parallel  to  H,  and  BC  (any  one  of 
four  different  lines),  whose  ^-projection  is  perpendicular  to  aHbH. 
Any  one  of  the  lines  BC  will  be  perpendicular  to  AB. 


FIG.  20. 


Proof. — If  AB  is  parallel  to  H,  any  plane  perpendicular  to 
AB  will  be  perpendicular  to  H,  and  will  therefore  be  a  projecting 
plane  on  H  of  all  lines  contained  in  it.  Let  the  student  finish 
the  reasoning  to  show  that  all  lines,  whose  /^-projections  are 
bHcH,  will  be  perpendicular  to  AB. 

Theorem  XXII. — Two  intersecting  lines  in  space  must  have 
then*  H-  and  V-projections  intersect  hi  points  hi  a  line  perpen- 
dicular to  GL.  Prove. 

Theorem  XXIII. — Two  intersecting  lines,  both  parallel  to  H, 
are  projected  on  H  hi  lines  parallel  and  equal  to  themselves,  and 
the  included  angle  will  be  equal  in  projection  to  the  angle  hi 
space.  Prove. 

Theorem  XXIV. — A  line,  that  is  parallel  to  an  oblique  plane, 
is  not,  in  general,  projected  parallel  to  the  traces.  The  exception 
is  only  possible  when  the  traces  are  both  parallel  or  both  per- 
pendicular to  GL, 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS     21 

Theorem  XXV. — A  line,  that  is  perpendicular  to  a  plane, 
always  has  its  projections  perpendicular  to  the  respective  traces 
of  the  plane. 

In  Fig.  21  let  BC  be  perpendicular  to  T  To  prove  that  bC 
is  perpendicular  to  tt,  the  //-trace  of  T. 

Let  fall  a  perpendicular,  Bb,  from  B  to  H.  The  lines  BC  and 
Bb  determine  a  plane  perpendicular  to  both  H  and  T,  and 
therefore  to  their  intersection,  tt.  As  the  plane  is  perpendicular 
to  tt,  its  intersection  bC  will  be;  that  is,  the  //-projection  is 


FIG.  21. 


perpendicular  to  the  //-trace, 
the  F-projection  and  F-trace. 


The  same  could  be  proved  for 


Theorem  XXVI. — Conversely,  a  line,  whose  projections  are 
perpendicular  to  the  respective  traces  of  a  plane,  is  perpendicular 
to  the  plane,  with  one  exception. 

Exception. — If  the  plane  is  parallel  to  GL,  the  line,  whose 
projections  are  perpendicular  to  its  traces,  may  not  be  perpen- 
dicular to  the  plane.  Prove  this  by  the  P-trace  of  such  a  plane, 
and  the  P-projection  of  such  a  line. 

Theorem  XXVII. — A  line  in  any  plane  pierces  H  hi  the  H- 
trace  of  the  plane,  unless  the  line  is  parallel  to  H.  Similarly,  it 
will  pierce  V  in  the  F-trace  of  the  plane  in  which  it  lies.  Prove. 

Theorem  XXVIII. — If  a  line  is  parallel  to  H,  it  is  parallel  to  the 
H-trace  of  any  plane  in  which  it  lies.  Similarly  for  F.  Prove. 

Theorem  XXIX. — Any  plane  that  is  passed  through  one  of 
two  parallel  lines  either  contains  or  is  parallel  to  the  other. 
Let  the  student  prove  this  with  a  figure.  Refer  to  Plane  and 
Solid  Geometry  theorems. 


22 


EXERCISES  IN  REVIEW 


20.  Directions. — All  the  exercises  here  given  can  be  drawn  in 
a  space  6  in.  X  7  in.,  with  GL  drawn  horizontally  across  the 
middle. 

Letter  all  points  carefully,  using  small  letters  for  all  projections, 
av,  bv,  etc.,  for  F-projections,  aH,  bH,  etc.,  for  H -projections,  ap, 
bp,  for  P-projections. 

Use  the  following  conventions  for  all  lines  etc. 

Given  lines — solid,  medium,  ^54  in. 

Required  lines — solid,  heavy  — ^^^^^^^•"~  ^2  i*1- 

Auxiliary  lines — solid,  light  — -  very  fine. 

Invisible  lines — dotted,  medium  — —  —  —  -«—  — —  ^4  in. 

Projectors — dotted,  light very  fine. 

Paths  of  Revolution — the  same. 

Traces  of  Given  planes — medium       — —   —   ^4  in. 

Traces  of  Required  planes — heavy  — —   —   — — •  ^3  m- 

Traces  of  Auxiliary  planes — light  -  very  fine. 

All  planes  are  considered  transparent,  and  there  will  be  no 
invisible  lines  until  problems  involving  solids  are  given.  ||  and 
J_  are  abbreviations  for  parallel  and  perpendicular. 

36.  Through  A(5  -  1  +-2)  draw  AB  ||  to  C(4  +  2  -  2)  D(2  +  1  -  M)- 

37.  Through    M(4  +  2  +  1)    draw    MN    ||    and    equal   to   O(3  -  1  -  2) 

P(l  +  M  ~  M). 

38.  From    A(4>£  -2  +  1)    draw    two    lines    intersecting    B(l  +  1  -  1) 
C  (3  +  3  -  2)  in  points  other  than  B  and  C. 

39.  Draw  two  lines  EF  and  KL  intersecting  in  the  point  D(3  —  1+1). 
Connect  these  two  lines  with  a  line  ||  to  H. 

40.  Draw  two  lines  MN  and  OP  intersecting  in  the  point  K(2M  +3  —  1). 
Connect  these  two  lines  with  a  line  ||  to  V. 

41.  Draw  a  line   J_  to  V  intersecting  A(l  +  M  -  2)  B(4  +  3  -  1)  and 
CUM  +2  -M)D(3M  +  1-1). 

42.  Draw  a  line  _|_  to  H  intersecting  E(1M  -3+2)  F(5  -  1  +  3)  and 
G(l>2'  -  1  +  2M)  K(4  -  2  +  1). 

43.  Draw    four   lines    at    right    angles    to    L(l+2-l)    M(3  +  M  -  1) 
through  the  point  M. 

44.  Through  N(2  -1+2)  draw  a  line  _L  to  T(l  -  3)  5(3  +  3). 

45.  Through  O(5  +  1  -  1)  draw  a  line  J_  to  S(5  +  3)  3(1  -  3). 

46.  Through  P(2,  0,  0)  draw  the  traces  of  a  plane  R  that  is  JL  to  A  (2  -  2  + 
2)  B(4  -  M  +  1). 

47.  Is  C(2  +  2  -  1)   D(2  +  M  -  2)   J_  to  the  plane  Q(  +  3)   «(  -  1)? 
If  not,  draw  the  traces  of  T,  that  is  _L  to  CD. 

Draw  the  P-projections  of  the  following  points  and  lines,  locating  P  4% 
in.  to  the  right: 

48.  A(l  +  2  +  1),  BUM  +  2  -  1),  C(2  -  1  -  2),  D(2M  -2  +  1),  and 
E(3,  0  +  1). 


DEFINITIONS,  NOTATIONS,  GEOMETRICAL  THEOREMS     23 

49.  The  lines  F(l  +  3  +  1)  G(l  -  1  +  1)  and  K(2  +  1  -  1)  G(2  -  2  - 
1).     What  relation  are  these  lines  to  H  and  Vf 

50.  M(l  +2—2)  N(l,  0  —  1).     What  angles  does  this  line  make  with  H 
and  Vf 

51.  Through  O(2  —  3  +  1)  draw  a  3-in.  line  ||  to  P,  45°  to  H.     In  what 
space  angle  does  the  line  terminate? 

52.  Let  A(3,  0  +  1^)  B(4^,  0  +  1)  be  the  diagonal  of  a -square  in  H. 
With  this  square  as  the  upper  base  of  a  cube,  draw  the  projections  of  a 
cube.     Draw  the  traces  of  the  six  planes  bounding  the  cube. 

VS 


N*>  X  -    ^  \>    -v 

X  /   *^'  ' 

-  /  ^  / 

XT  R.^'  <*!  w/ 

.X"  .X^ 


FIG.  22. 

53.  Draw  the  three  projections  of  c,  hexagonal  prism  in  /,  its  center  line  || 
to  H  and  V,  with  a  1^-in.  base  and  2-in.  altitude.     Draw  the  H-  and 
F-traces  of  the  plane  of  one  of  its  oblique  faces. 

54.  Draw  the  prism  given  in  Ex.  53  in  ///,  and  draw  the  H-  and  F-traces 
of  the  plane  of  one  of  its  oblique  faces. 

55.  Let  C(2,  0  —  1)  D(3K>  0—2)  be  the  base-diagonal  of  a  hexagonal 
prism  of  2-in.  altitudj,  standing  on  H.     Draw  the  traces  of  the  six  face 
planes  of  the  prism. 

56.  Write  the  notation  of  the  planes  in  Fig.  22,  and  describe  them.     Use 
your  own  dimensions. 


INTRODUCTORY 

21.  In  the  solution  of  problems  that  compose  this  chapter, 
we  (1)  analyze  the  operation  in  space,  (2)  offer  proof  of  the  correct- 
ness of  the  analysis,  when  it  seems  necessary,  and  (3)  show  the 
construction  of  a  typical  example  of  the  problem  "on  the  flat." 
The  analysis  should  be  learned,  not  by  memorizing,  but  by 
visualizing  the  situation  and  process  in  space. 

Descriptive  Geometry,  as  the  science  of  drafting,  is  intended 
to  solve  drafting  difficulties.  Drafting  presents  very  few  diffi- 
culties when  objects  are  in  their  " natural  position;"  that  is, 
when  the  lines,  plane  figures,  and  solids  are  parallel  to  H  or  V. 
It  is  only  when  the  objects  are  in  oblique  positions,  that  we  find 
their  relations  difficult.  These  difficulties  show  themselves  in 
distortions,  called  "foreshortening." 

The  obvious  general  solution,  therefore,  is  to  bring  the  objects 
into  their  simplest  relations  (parallelism),  and  this  is  done  by 
one  of  two  methods.  )The  first  method  might  be  called  the 
"draftsman's  method,"  and  is  equivalent  to  the  changing  of 
position  by  the  observer  in  order  to  bring  the  objects  into  their 
desired  relation.  The  second  might  be  termed  the  "mathe- 
matician's method,"  and  consists  in  rotating  the  objects,  or  in 
securing  the  results  by  means  of  a  series  of  operations,  such  as 
passing  planes,  making  intersections,  rotating,  and  the  like,  which 
finally  achieve  the  solution.  '  Both  methods  have  their  advantages, 
and  both  are  presented  in  many  cases,  but  where  the  superiority 
of  either  is  conspicuous,  that  method  only  is  offered. 

It  has  been  shown  that  certain  situations  can  only  be  repre- 
sented by  the  use  of  a  third  plane  of  projection,  the  Profile,  per- 
pendicular to  both  H  and  V.  In  a  like  manner,  there  are  certain 
solutions  of  this  type,  but  different  in  situation,  that  can  best  be 
represented  by  projecting  on  a  plane  that  is  perpendicular  to 
H  or  V,  but  not  to  both.  This  method,  the  draftsman's  method, 
is  merely  a  variant  on  profile  projection,  and  is  performed  in  the 

24 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     25 

same  way.     It  requires  no  analysis,  hence  only  the  construction 
is  given,  when  it  is  employed. 


FUNDAMENTAL  PROBLEMS 

22.  There  are  six  problems  that  depend  on  no  other  previous 
problems  for  their  solution  and  are  themselves  necessary  to  the 
solution  of  succeeding  problems.  To  the  complete  mastery  of 
these,  then,  it  is  evident  that  success  is  due.  These  fundamentals 
are: 

Problem  I.  To  project  a  line  on  an  auxiliary  plane  perpendicular  to  H 
or  V. 

Problem  II.     To  find  the  trace  of  an  oblique  plane  on  an  auxiliary  plane. 
Problem  III.     To  find  the  H-  and  F-piercing  points  of  a  line. 
Problem  IX.     To  revolve  a  point  about  an  axis  perpendicular  to  H  or  V. 
Problem  XII.     To  revolve  a  point  about  a  line  in  H  or  V  into  that  plane. 
Problem  XVII.     To  find  the  intersection  of  two  planes. 


AUXILIARY  PLANES  OF  PROJECTION 


FIG.  23. — Showing  H,  V,  and  Q  planes. 

23.  By  projecting  an  oblique  line  on  a  plane  parallel  to  it  and 
perpendicular  to  H,  the  true  length  of  the  line  and  its  true  angle 
with  H  are  shown  by  this  projection,  or  if  the  plane  were  perpen- 
dicular to  V,  its  true  relation  to  V  is  shown  in  the  same  way. 
We  designate  these  planes  as  follows: 

Q  is  the  plane  perpendicular  to  H  and  oblique  to  V. 
R  is  the  plane  perpendicular  to  V  and  oblique  to  H. 


26 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


If  a  second  removal  is  required : 

X  is  the  plane  perpendicular  to  Q  and  oblique  to  H. 
Y  is  the  plane  perpendicular  to  R  and  oblique  to  V. 

The  accompanying  figure  shows  the  projection  of  a  line  on 
Q;  note  that  the  Q-projections  of  the  two  points,  A  and  B,  are 
the  same  in  elevation  as  their  F-projections.  This  fact  is  the 
chief  instrument. 

FUNDAMENTAL  PROBLEM  I 

24.  Problem  1. — To  project  an  oblique  line  on  an  auxiliary 
plane  parallel  to  the  given  line,  and  perpendicular  to  H  or  V. 
Let  AB  (Fig.  24)  be  the  given  line. 


FIG.  24. 

Construction. — 1.  Draw  HQ  parallel  to  aHbH.     (VQ  is  drawn 
here,  but  is  not  necessary  in  this  problem.) 

2.  Run  projectors  from  aH  and  bH  perpendicular  to  HQ. 

3.  Locate  aQ  on  its  projector  the  same  distance  from  HQ  that 
av  is  from  GL.    Locate  bQ  similarly. 

Conclusion. — The   line  aQbQ  is  the   projection  of  AB  on  Q. 
(Let  the  student  give  the  full  explanation.) 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     27 

Corollary  I. — The  Q-projection  of  AB  is  the  true  length  of  the 
line.  Why? 

Corollary  II. — The  angle  between  aQbQ  and  HQ  is  the  true 
angle  between  AB  and  H.  Why? 

Note. — When  the  other  trace  of  Q  or  R  is  not  needed,  it  will  be  omitted 
from  figures  hereafter. 

25.  Supplementary  Problem. — To  lay  off  a  given  distance  on  a 
given  line  from  a  given  point. 

,bv 


FIG.  25. 

Required. — To  lay  off  1  in.  on  AB  (Fig.  25)  from  A  toward  B. 
Constructions. — 1.  Obtain  the  Q-projection  of  AB. 

2.  Lay  off  aQcQ  equal  to  1  in. 

3.  Project  CQ  back  to  CH,  thence  to  cv. 
Conclusion. — AC  is  1  in.  long. 

EXERCISES 

26.  Note. — Exercises  throughout  this  book  are  presented  in  two  ways, 
with  dimensioned  coordinates,  and  graphically.  The  various  sets  are  not 
exact  duplicates.  The  graphic  "layouts"  are  not  always  given  dimensions, 
that  being  left  to  the  student  or  instructor. 

Project  on  Q  the  following  lines  and  measure  their  true  length  and  their 
angle  with  the  horizontal. 

57.  A(2  +  2  -  1)  B(4  -  K  -  2). 

58.  C(l  +  1  +  1)  D(2  +  3  +  3). 

59.  E(5  +  1  +  2)  F(3  -  3  +  Yz). 

60.  G(2  -  1  -  2)  K(4  -  2  -  1). 

61.  L(l  -3  +  1)  M(3  -  \YZ  +  3) 

62.  O(4  +  2  -  3)  P(2  +  1  +  2). 


28 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Project  on  R  the  following  lines,  and  measure  their  true  lengths  and  inclina- 
tion from  the  vertical. 

63.  A(4  +  2  +  1)  B(2  +  1  -  2). 

64.  C(l  -  1  -  2)  D(3  -  2K  -  1). 

65.  E(2  -  1  +  2)  F(3  -  2%  +  1). 

66.  G(3  -  3  -  3)  K(5  -  1  -  1). 

67.  L(2  +  2  -  H)  M(3M  +  1-3). 

68.  N(2  +  1  -  1)  O(4  -  1  -  3). 

Lay  off  on  the  above  lines  a  distance  of in.  (amount  specified  by  in- 
structor) from  the  first  point  toward  the  second. 


v 

Cfc" 


f¥ 


6  OM 


Graphic  Layout  No.  1. — Make  Q-projections  of  these  lines.     Dimensions 
to  suit. 


10 


12 


Graphic  Layout  No.  2. — (a)  Make  R-projections  of  these  lines.  (6)  Lay 

off  on  the  above  lines  a  distance  of in.,  from  the  first  point  toward  the 

second. 

Note. — The  requirements  for  these  layouts  may  be  transposed,  if  desired. 

FUNDAMENTAL  PROBLEM  II 

27.  Problem  2. — To  find  the  Q-trace  or  R-trace  of  a  plane 
oblique  to  H  and  V. 

Required,  to  find  the  Q-trace  of  the  plane  T  in  Fig.  26. 

Construction. — 1.  Draw  HQ  perpendicular  to  HT,  and  VQ 
perpendicular  to  GL. 

2.  Construct  a  right  triangle  on  HQ  as  a  base,  making  the 
altitude  equal  to  that  part  of  VQ  included  between  GL  and  VT. 

Note. — This  is  equivalent  to  revolving  into  H  the  right  triangle  cut  by  Q 
from  H,  V  and  T. 

Conclusion. — The  hypothenuse  of  this  triangle  is  the  Q-trace 
of  T,  hereafter  written  QT. 

Note. — Thus  T  is  a  projecting  plane  on  Q,  and  all  points  in  T  are  projected 
on  Q  in  QT.  This  makes  it  a  simple  instrument  for  all  problems  relating 
to  points  in  T. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     29 

28.  Supplementary  Problems. — To  find  the   dihedral  angle 
between  an  oblique  plane  and  a  plane  of  projection. 


X 


FIG.  26. 

Conclusion. — The  angle  6  in  Fig.  26  is  the  true  measure  of  the 
angle  between  T  and  H.  If  the  R-trace  of  T  had  been  found,  it 
would  have  yielded  the  angle  between  T  and  V. 

Proof. — The  plane  Q  is  perpendicular  to  the  intersection  of  T 
and  H,  therefore  the  angle  between  HQ  and  QT  is  the  measure 
of  the  dihedral  angle.  R  does  the  same  for  T  and  V. 


29. 


EXERCISES 


X'v 


Graphic  Layout  No.  3. — Find  the  Q-  and  R-traces  of  plane 


.     Also 
measure  the  dihedral  angles  between  the  planes  and  H  and  V,  in  degrees. 

Find  the  Q-trace  and  R-trace  of  the  various  planes  given  here.  Also 
measure  the  dihedral  angle  in  degrees  that  each  of  these  planes  makes  with 
H  and  V. 

69.  8(1  +  3)  4(1  -  2). 

70.  T(5  -  1)  2(5  +  2). 


30 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


71.  U(2  +  2)3^(5  -2). 

72.  W(3  -  1)  5(3  -  3). 

73.  Z(4  +  3)  iy2( 

74.  S(l  +  3)  (5  +  2)  (1  - 

75.  T(5  -  3)  (1  -  2)  (5  + 

76.  U(l  +  3)  2^(5  -  1). 

77.  W(l  -  1)  3(5  + 


(5  - 
(1  +  1). 


FUNDAMENTAL  PROBLEM  III 

30.  Problem  3.  —  To  find  the  points  in  which  a  line  pierces  H 
and  V. 

Analysis.  —  1.  Extend  the  F-projection  of  the  line  until  it  in- 
tersects GL,  and  project  this  point  to  the  //-projection.  The 
point  thus  found  will  be  the  //-piercing  point. 


FIG.  27. 

Interchanging  V  and  H  in  the  analysis  will  yield  the  F-piercing 
point. 

Proof. — All  points  in  H  have  their  F-projections  in  GL,  hence 
this  point  is  in  H,  and  therefore  where  the  line  pierces  H. 

Construction. — Let  AB  (Fig.  27)  be  any  oblique  line. 

1.  Extend  avbv  to  GL. 

2.  Project  the  intersection  to    El  on  aHbH,   or  its  extension. 
(Q  is  the  symbol  hereafter  used  for  the  H  -piercing  point.) 

3.  Extend  aHbH  to  GL. 

4.  Project  the  intersection  to    O  on  avbv,  or   its   extension. 
(O  is  the  symbol  for  F-piercing  point.) 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     31 


Axioms.  —  A  line  parallel  to  V  has  no  0,  hence  the  ^-projection 
of  such  a  line  is  parallel  to  GL. 

A  line  parallel  to  H  has  no  Q,  hence  the  F-projection  of  such 
a  line  is  parallel  to  GL. 


A 

/  i 

rr/       1                              1 

T     i                \ 

a" 


\\ 


1 


I 

*-- 


FIG.  28. — H-  and  F-piercing  points  of  lines  located  in  P-planes. 

Construction. — An  examination  of  Figs.  28  and  28o  will  be 
sufficient  to  show  the  operation  in  this  case. 


(^ 


31.  Supplementary  Problem. — To  find  the  point  where  a  line 
pierces  a  profile  plane,  and  show  it  in  P-projection. 


32 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Construction. — Let  AB  (Fig.  28b)  be  the  given  line.     Required 
its  P-piercing  point.     Locate  P  at  any  suitable  point. 


C  is  the  piercing  point. 


FIG.  286. 

1.  Extend  avbv  to  the  P-trace  at  cv. 

2.  Extend  aHbH  to  the  P-trace  at  CH. 

3.  Revolve  P  into  V,  locating  cp. 

EXERCISES 

32.  Locate  the  H-  and  F-piercing  points  of  the  following  lines,  and  tell 
what  angles  they  traverse.  If  P  is  required,  place  it  at  4K  in.  from  the 
margin. 

78.  A(I  +  2  -  K)  B(2K  +y2  -  2K),  c(3K  +  K  +  2)  D(5  +  IK  + 
K). 

79.  E(l  -  K  -  1M)  F(3  -  2  -  K),  G(3K  +  2  +  1)  K(5  -  1  -  2). 

80.  M(l  +  2  +  2)  N(2K  -  1  -  1),  0(3  -  1  +  2)  P(5  -  2  +  K). 

81.  A(l  +  2  +  1)  B(3  +  K  -  2),  C(3  +  2  +  1)  D(4  +  1  +  K). 

82.  E(2  -  K  -  IK)  F(3  -  IK  -  2),  G(3  -  1  +  2)  K(5  -  3  -  1). 

83.  M(l  +  2  -  IK)  N(l  +  1  -  K),  0(2  -  2  -  2^)  P(2  -  1  -  K). 

84.  A(l  +  2  +  K)  B(l  +  %  +  IK),  C(2  -  2  +  K)  D(2  -  1  -  K). 

85.  E(l  -  1  -  2)  F(l  +  2K  +  K),  G(2  -  2K  -  M)  K(2  +  K  +  1). 

86.  M(l  +  2  -  2)  N(l  -  1  -  K),  0(2  +  2  +  IK)  P(2  -  IK  +  M)- 

87.  A(l  +  2  -  K)  B(l  -  K  -  1),  C(2  +  K  -  2)  D(2  -  2  +  K). 

88.  E(l  +  2  -  1)  F(l  -  1  +  IK),  G(2  -  2  -  1)  K(2  +  1  +  1). 

Find  the  P-piercing  points  of  the  following  lines,  locating  P  4K  in-  to  tne 
right. 

89.  A(2  +  K  +  2)  B(3K  -  1  +  K). 

90.  C(2  +  IK  -  1)  D(3K  +  2K  -  K). 

91.  E(1K  +  1  +  IK)  F(3K  +  1  +  IK)- 

92.  G(2  +  K  -  1)  K(3K  +  IK  -  1). 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     33 


93.  M(1K  -  2  +  M)  N(4  -  1  +  2). 

94.  O(2  -  3  +  2)  P(4  -  1  +  1). 

95.  A (2  +  3  -  2)  B(4  +  1  -  1). 


ov       *> 

t !°" 


\\fv        fr,H     ipM    jav 

V    t  i»  lb" 


0)  (2)         (3)         (4)          (5)    (6)  (7)   (8)  (9)  (10) 

Graphic  Layout  No.  4. — Find  the  H  and  O  of  the  lines. 


mr 


0) 


(2)  (3) 


(4) 


(51  (6) 

Graphic  Layout  No.  6.  —  Find  the  P-piercing  points  of  the  lines.     Place 
P  \%  in.  to  right  of  margin. 

LOCATING  POINTS  AND  LINES  IN  PLANES 

33.  Problem  4.  —  To  assume  a  line  in  a  plane. 
Analysis.  —  1.  Assume  a  point  in  the  F-trace. 


FIG.  29. 


2.  Assume  a  point  in  the  H-trace. 

3.  Connect  the  respective  projections  of  the  two  points. 
Conclusion. — The  line  thus  found  will  lie  entirely  in  the  plane, 

because  two  of  its  points  lie  in  the  plane. 


34  PRACTICAL  DESCRIPTIVE  GEOMETRY 

Construction. — Let  T  (Fig.  29)  be  the  plane. 

1.  Assume  av  in  VT;  aH  will  be  in  GL. 

2.  Assume  bH  in  HT;  bv  will  be  in  GL. 

3.  Draw  avbv  and  aHbH.     AB  will  be  entirely  in  T. 

Special  Case. — To  assume  a  line  parallel  to  H  or  V,  lying  in 
an  oblique  plane. 

If  parallel  to  H,  draw  the  F-projection  parallel  to  GL  from  an 
assumed  point  in  the  F-trace.  From  the  //-projection  of  this 
point  draw  the  //-projection  parallel  to  the  //-trace. 

In  Fig.  29a,  the  line  CD  is  a  parallel  line  to  H  lying  in  S,  and 
the  line  EF  is  parallel  to  V  lying  in  R. 


FIG.  29a. 

34.  Problem  5. — Given  one  projection  of  a  point  in  a  plane, 
to  locate  the  other  projection. 

Construction. — Through  the  given  projection  draw  the  pro- 
jection of  a  line  and  assume  the  line  to  be  in  the  plane,  thus 
locating  the  other  projection.  The  required  projection  of  the 
point  will  be  on  the  second  projection  of  the  assumed  line. 

Hint. — The  convenient  line  for  most  cases  of  the  foregoing  is  a 
line  parallel  to  H  or  V. 

Note. — If  the  given  projection  of  the  point  lies  in  the  projecting  trace  of  a 
projecting  plane,  the  point  is  indeterminate,  and  can  be  one  of  an  infinite 
number  of  points  lying  in  the  plane. 

Problem  6. — To  find  the  traces  of  a  plane  determined  by 
three  given  points. 

Analysis. — 1.  Join  the  points,  two  and  two,  by  lines. 

2.  Find  the  piercing  points  of  the  lines. 

3.  Draw  the  H-trace  through  the  //-piercing  points,  and  the 
F-trace  through  the  F-piercing  points. 

Proof. — The  lines  have  their  piercing  points  in  the  respective 
traces  of  the  plane,  therefore  the  lines  (and  all  their  points)  lie 
in  the  plane  thus  determined. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     35 

Construction. — Let  A,  B,  and  C  be  the  given  points  (Fig.  30). 

1.  Join  the  projections  of  AB  and  AC. 

2.  Find  the  four  piercing  points  of  AB  and  AC. 

3.  Through  the  //-piercing  points  draw  HT,  and  through  the 
Tr-piercing  points  draw  VT. 

Check. — 1.  The  line  BC  should  have  its  piercing  points  in  the 
respective  traces  of  T,  and 

2.  The  traces  should  meet  in  a  point  on  GL,  or  be  parallel 
toGL. 


FIG.  30. 

Hints. — 1.  Sometimes  the  points  may  be  so  placed  that  the 
piercing  points  are  not  all  to  be  found  within  the  limits  of  the 
problem.  In  such  a  case,  an  auxiliary  line  may  be  drawn  from 
any  of  the  three  points  to  any  point  on  the  line  joining  the  other 
two.  For  instance,  if  such  a  proceeding  were  necessary,  a  line 
could  be  drawn  from  B  to  any  point,  M,  on  the  line  AC.  BM 
would  lie  in  the  required  plane,  and  its  piercing  points  would 
help  to  determine  the  traces. 

2.  Ordinarily,  only  three  piercing  points  are  necessary  to  be 
found,  because  when  the  fl-trace  is  found,  its  intersection  with 
GL  may  be  joined  to  one  of  the  F-piercing  points. 

If  two  intersecting  lines  are  given. — This  makes  no  change  in 
the  problem. 

If  two  parallel  lines  are  given. — Find  their  piercing  points 
and  draw  the  traces,  as  in  the  previous  case. 

If  a  point  and  a  line  are  given. — The  point  may  be  connected 
with  any  point  in  the  line,  thus  making  two  intersecting  lines, 
or  through  the  point  a  line  may  be  drawn  parallel  to  the  given 
line,  thus  making  two  parallel  lines. 


36 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


EXERCISES 

35.  Locate  the  following  lines  in  the  planes  given.     The  points  to  be 
determined  are  lettered  as  unknowns,  x,  y,  z,  etc.,  without  +  or  —  signs. 
1   96.  AB,  any  oblique  line  in  T(l  +  3)  1(4  -  2). 

97.  CD,  any  oblique  line  in  8(5  +  2)  2(2  -  3). 

98.  EF,  any  oblique  line  in  any  plane  oblique  to  GL. 

99.  Draw  a  line  in  the  third  angle  that  lies  in  T(l  +  2)  3(1  -  3). 

100.  Locate  C(2  +  1,  x)  D(3,  +  2,  y)  in  8(1  +  2)  4(1  -  2^).     State  values 
of  x  and  y. 

101.  Locate  E(4  +,  x)  F(5  +  2,  y)  in  W(l  +  3)  3(5  -  1). 

102.  Draw  a  line  in  8(1  +  2)  4(1  -  2^)  that  is  ||  to  H,  1  in.  above  H. 

103.  Draw  a  line  in  W(l  +  3)  3(5  -  1)  that  is  ||  to  F,  H  in.  behind  F. 

104.  Locate  M(5  +  1,  x)  N(3,  0,  y)  in  W(l  +  3)  3(5  -  1). 

105.  Locate  O(3  +  2,  x)  P(4  +  %,  y)  in  T(l  +  3)  3(5  -  3). 


4      VU 


VW 


"  ' 


10 


HX 


Graphic  Layout  No.  6.— Locate  the  other  projections  of  the  line  in  the 
given  plane . 

106.  Find  the  locus  of  all  points  in  S(5  +  2)  2(5  -  3): 
(a)  1  in.  above  H. 

(6)   1  in.  below  H. 

(c)  2  in.  behind  V. 

(d)  2  in.  before  V. 

107.  Locate  in  R(  +  2)  <*>  (  -  1^)  the  following  lines: 
A(l  +  1,  x)  B(2  +  iy2,  y). 

C(2,  x,    -  1)  D(4,  y  -  2K). 

EF,  the  locus  of  all  points  1  in.  before  V. 

108.  Locate  the  following  lines  mX(  +  lK)°°(-l): 
F(l  -  1,  x)  G.(2  -  2,  y). 

K(2K  +  2^,  x)  L(4  +,  y). 

MN,  the  locus  of  all  points  %  in.  above  H. 

109.  Locate  O(2  +  2,  x)  P(4  +  1,  y)  in  T(l  +  3)  (5  +  1H)  d  -  2)  (5  -  1). 
Locate  the  following  points  in  the  planes  given : 

110.  A(l,  x  -  1),  B(3  +  3,  y)  in  8(1  +  3)  5(1  -  \Y2). 

111.  C(1K  -  2,  x),  D(4,  y  +  2)  in  U(5  -  1)  1(3  +  3). 

112.  E(2  +  1,  x),  F(3,  y  -  3)  in  W(  +  2)  co  (  -  1). 

113.  G(2  +  IK,  x),  K(3  +  3,  y)  in  Q(  +  2)  =0  (  +  1). 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     37 


114.  L(l  +  1,  x),  M(4^,  y  +  1)  in  R(l  +  2)  3(5  -  2). 

115.  N(l,  x  +  K),  0(3  +  2,  y)  in  S(l  +  3)  5(1  +  1). 

116.  P(l  +  iy2,  x),    A(4,  y  -  ^),    B(4H  +  2,    z)  in    T(l  +  3)  (5  + 
(1  -2-)  (5  -  1). 


3      H 
.ev~ 


VQ 


.m 


Graphic  Layout  No.  7.  —  Given  one  projection  of  points  in  planes,  to 
locate  the  other. 


Draw  the  traces  of  the  planes  containing  the  following  points,  lines,  and 
plane  figures: 

117.  A(2  +  2  -  1),  B(3  +  K  -  3),  and  C(3K  +  1^  -  IK)- 

118.  D(2  +  1  -  K),  E(2K  -Yz+Yz)  and  F(4  +  11A  +  YA- 

119.  G(2  -  2  +  1),  K(3  -  K  +  3),  and  L(3K  -  IK  +  IK). 

120.  M(2J^  +  K  -  K)  N(3K  +  IK  -  %),  and  O(4>i  +  K  -  1M). 

121.  A(l  -  1  +  2K)  B(4  -  M  +  1),  and  C(3  -  2  +  K). 

122.  The  triangle  D(1K  +  1  -  1H)  E(3  +  1  +  K)  F(4  -  1  +  2). 

123.  The    parallelogram    M(1K  -  1+  M)    N(2  -  1%  +  %)    O(2^  - 
%  +  1%)  P(3M  -  1  +  1%). 

124.  HT  is  a  line  through  A(4,  0,  0)  and  B(2,  0  -  2K).     T  contains 
D(l  +  1  -  1H).     Draw  VT. 


lb 


la 


6    a 


a 


ok 


Graphic  Layout  No.  8. — Locate  these  points  in  either  /  or  777,  and  draw 
the  traces  of  the  planes  containing  them. 


38 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


125.  The  points  D(2^  +  1M  -  D,  E(3^  +%-!),  F(3>*  +  2>£  -  1), 
and  G(4  +  1^  —  1)  are  the  corners  of  a  vertical  base  of  a  cube.     Draw 
the  projections  of  the  cube  and  the  traces  of  its  six  bounding  planes. 

126.  L(3,  0  —  1)  M(4,  0  —  2^)  is  the  base  diagonal  of  a  square  pyramid  of 
2-in.  altitude,  standing  on  H.     Draw  the  traces  of  the  four  oblique 
planes  forming  its  faces. 

127.  O(3,  0  —  1M)  is  the  center  of  the  base  of  a  2-in.  hexagonal  oblique 
pyramid,  whose  apex  is  P(2  +2  —  1).     Hexagon  base  is  in  H;  draw 
the  traces  of  the  six  oblique  bounding  planes. 

36.  Problem  7. — To  pass  a  plane  through  a  given  point, 
parallel  to  two  given  lines. 

Analysis. — 1.  Draw  through  the  given  point  two  lines,  parallel 
respectively  to  the  given  lines. 

2.  Pass  a  plane  through  the  lines  thus  drawn. 

Conclusion. — This  plane  contains  the  given  point  and  is 
parallel  to  both  lines,  from  the  geometrical  theorem  that  a  line, 


external  to  a  plane,  that  is  parallel  to  any  line  in  that  plane,  is 
parallel  to  that  plane. 

Construction. — Let  A  (Fig.  31)  be  the  given  point,  and  BC 
and  EF  the  given  lines. 

1.  Through  A  draw  MN  ||  to  BC,  and  OP  ||  to  EF. 

2.  Find  the  H-  and  F-piercing  points  of  MN  and  OP. 

3.  Through  these  piercing  points  draw  the  traces  of  the  plane. 

Note. — If  the  given  lines  are  parallel,  the  required  plane  is  indefinite,  and 
an  infinite  number  of  planes  may  be  passed. 

37.  Problem  8. — To  pass  a  plane  through  a  given  line  parallel 
to  another  given  line. 

Analysis. — 1.  Through  any  point  of  the  given  line,  draw  a 
line  parallel  to  the  second  line. 

2.  Pass  the  required  plane  through  the  two  intersecting  lines. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     39 


Let  the  student  make  the  construction. 

Note.  —  If  a  plane  be  passed  through  the  second  line  parallel  to  the  first, 
the  two  planes  will  be  parallel. 

EXERCISES 

38.  Pass  planes  through  the  following  points  parallel  to  the  lines  given  : 

128.  Through    O(3  -  1  +  1)    ||    to    A(l  -  1^  +  1)    B(2  -  M  +  2)    and 
C(4  -  1  +  i)  D(5  -  1M  +  2). 

129.  Through    A(4  +  1  -  1)     ||    to    E(l  +  2  -  ^)    F(l  +  1  -  1)    and 
G(2  +  Y2  -  1)  K(3  +2-2). 

>130.  Through  M(4  +  1  -  1^)  ||  to  L(l^  +  1-1)  N(2^,  0  -  K)  and 
0(3  +  1  +  K)  P(4  +  2  +  iy2). 


k"o" 


v    \ 


Graphic  Layout  No.  9. — Through  the  given  point,  pass  a  plane  parallel 
to  the  two  given  lines. 

131.  Through  A(l  +  3  -  2)  B(2  +  1  +  1)  pass  a  plane  ||  to  C(3  +  1  -1) 

D(4  +  iy2  -  2). 

132.  Through  E(4  -  1  +  2)  F(3  -  2  +  1)  pass  a  plane  ||  to  G(l  +  1^  -  2) 
K(2  +  1  -  1). 

133.  Through  M(3  -  2  +  1)  N(4  -  K  -  2)  pass  a  plane  ||  to  O(l  +  2  +  1) 
P(2  +  3  -  1). 


3m' 


A 

v      M  /  I 

.«  °r   i 


W 

*     c 


r  3 


-J!H     ikM 
laH 


Graphic  Layout  No.  10. — Through   the  line  -  — ,    Ex. ,    pass  a 

plane  parallel  to  line . 

REVOLUTION  OF  POINTS 

39.  Definition. — The  revolution  of  a  point  about  a  line  (called 
its  axis)  consists  in  moving  the  point  in  a  plane  perpendicular  to 
the  axis,  and  keeping  it  at  a  constant  distance  from  the  axis. 

The  Path  of  Revolution  is  a  circle. 


40 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


FOURTH  FUNDAMENTAL  PROBLEM 

40.  Problem  9. — To  revolve  a  point  through  any  given  angle 
about  an  axis  perpendicular  to  H  or  V. 

L.     Construction. — Let  O  (Fig.  32)  be  the  point,  AB  the  axis  per- 
pendicular to  H,  and  6  the  given  angle. 

1.  With  aHbH  as  a  center  draw  a  circle  through  OH. 

uV  2.  Through  ov  draw  a  line  parallel  to  GL. 

(These  will  be  the  H-  and  F-projections  of 
the  path  of  revolution.) 
°"         3.  Move  OH  through  the  given  angle  to 


---i- 


4.  Project  to  Oiv. 

Conclusion. — This  gives  the  new  posi- 
tion of  the  point.  If  the  axis  were  perpen- 
dicular to  V,  the  circle  would  be  projected 
on  V  in  its  true  size  and  on  H  as  a  line  par- 
allel to  GL. 

41.  Problem  10. — To  revolve  a  line  about 
a  vertical  or  horizontal  axis  through  any 
given  angle. 

Analysis. — 1.  As  the  line  is  the  sum  of 
all  points  in  it,  take  any  two  points  of  it 
and  revolve  them  about  the  axis,  as  in  Problem  9. 

2.  Join  these  revolved  points,  and  the  resulting  projections 
will  be  projections  of  the  line. 


FIG.  32. 


42. 


EXERCISES 


134.  Revolve    A  (4  +  1  -  1)    about  B(3,  0-2)   C(3  +  3  -  2)  counter- 
clockwise 120°. 

135.  Revolve  D(4  +  2  -  1)  about  E(3  +  }£,  0)  F(3  +  K  -  2)  clockwise 
240°. 

136.  Revolve  G(l  +  1  -  1)  about  K(2  -  1  +  1)  L(2  -  1  +  2)  clockwise 
135°,  and  M (5  -  1  +  1)  about  N(4,  0  +  2^)  P(4  -  2  +  2^)  until 
it  reaches  H. 

137.  Revolve  A (3  -  1  -  H)  clockwise  90° about  B(4,  0  -  1)  C(4  -  2  -  1). 
Continue  the  revolution  until  it  returns  to  the  fourth  angle. 

138.  Revolve   D(5  -  Y2  -  1)  counter-clockwise   90°  about   E(4  +  K,  0) 
F(4  +  K  —  2).     Continue  the  revolution  until  it  reaches  H  again. 

139.  Revolve  G(4  +  M  -  1)  K(3  +  2  -  2)  about  P(3  +  1,  0)  O(3  +  1  -  3) 
clockwise  135°. 

140.  Revolve   L(2  +  2  -  1)    M(3^  +  1  -  2^)    about    K(l^,    0-2) 

+  3  —  2)  into  a  position  parallel  to  V. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     41 

141.  Revolve  A(l  -  1  +  2)  B(3  -  2  +  H)  clockwise  90°  about  C(2,  0  +  1) 
D(2  -  2  +  1). 

142.  Revolve  E(l  +  1  +  1)  F(2^  -1+2)  clockwise  75°  about  G(2  -  2  + 
2)  K(2  +  2  +  2).     Also  revolve  it  ||  to  V. 


\ 

bv 

'd 

5 

4 

5 

V 

V 

tm 

1 

c 

rd 

V 

e 

!*" 

V                        I  V 

q                  k 

H 

«'  i 

1 

1 
1 

V 

T° 
1 

1^   i 

m             n 

i  T°V    ! 

1 

CM 

ev 

l 

1    '       1 

H 

cU 

bH    AmH 

in* 

I 
Jof 

v  : 

d" 

fV 

g«              kn 

OH 

Graphic  Layout  No.  11. — Revolve  the  given  point 
about  the  given  axis. 


clockwise 


Ofb 


/v 


elf" 


I        I    I 


Graphic    Layout    No.   12. — Revolve    given  line 
wise  about  axis . 


clock- 


43.  Problem  11. — ^To  find  the  true  length  of  a  line  in  space,  by 
revolution. 

Discussion. — To  find  the  true  length  of  a  line  by  projecting  it 
on  an  auxiliary  plane  is  solved  in  Problem  I,  Corollary  I.  Another 
method  in  quite  as  much  use  is  that  of  revolving  the  line  parallel 
to  one  of  the  planes  of  projection,  using  the  principle  of  Problems 
9  and  10. 

Analysis. — 1.  Assume  an  axis  through  the  given  line  perpen- 
dicular to  H  or  V. 

2.  Revolve  the  line  as  in  Problem  10  until  it  becomes  parallel 
to  the  plane  to  which  the  axis  is  parallel. 

Conclusion. — The  projection  on  this  plane  is  the  true  length  of 
the  line. 

Proof. — The  line  remains  unchanged  in  revolving  about  its 
axis,  because  every  point  in  the  line  moves  through  the  same  angle 


42 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


in  a  series  of  parallel  circles.  This  means  that  every  point  re- 
mains in  the  same  relation  to  the  axis  as  every  other  point 
throughout  the  revolution,  and  therefore  all  points  in  the  line 
remain  in  the  same  relation  to  each  other,  and  the  line  is 
unchanged. 

Note. — In  revolving  parallel  to  V,  the  //-projection  is  not  changed  in 
length,  but  the  F-projection  is. 

Construction. — Fully  shown  in  Fig.  33. 


FIG.  33. 


44.  The  true  length  of  a  line  in  a  profile  plane. 

The  P-projection  shows  its  true  length. 


45. 


EXERCISES 


143.  Measure  A(l  +  1  -  1)  B(2  +  3  -  2)  by  revolving  ||  to  H,   (2)  ||  to 
V.     Check  result  by  Problem  1,  Corollary  1. 

144.  Measure  C(5  -  1  +  3)  D(3  -3+2),  (1)  by  revolving  ||  to  V,  (2)  || 
to  H.     Check. 

145.  Measure  E(2  +  1  +  1)  F(4  +  2^  +  2^),  (1)  by  revolving  ||  to  V, 
(2)  |1  to  H.     Check. 

146.  Measure  G(3  -  3  -  3)  K(l  +  1-1),   (1)  by  revolving  ||  to  H,  (2) 
||  to  V.     Check. 

147.  Measure  L(2  +  1  -  2)  M(3  -1+1),   (1)  by  revolving  ||  to  V,  (2) 
||  to  V.     Check. 

148.  Measure  M(4  -  2  +  1)  N(5  +  1  +  1^),  (1)  by  revolving  ||  to  H,  (2) 
||  to  H.     Check. 

Measure  the  following  lines.     Place  Pat  4^. 

149.  A(l  -  2  +  M)  B(l  -  Y2  +  1)  and  C(2  -  1  -  2)  D(2  -  2  -  Yz)- 

150.  E(l  +  3  -  1)  F(l  +  1  -  2)  and  G(2  +  ^  +  W±)  K(2  +  3  +  Y*)- 

151.  M(l  +  2  -  1)  N(l  -Yz  -2)  and  O(2  -  1  +  1)  P(2  +  2  +  H)- 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     43 

152.  A(l  -  2  +  1)  B(l  +  2  -  2)  and  C(2  +  1  +  2)  D(2  -  2  -  ^). 

153.  E(l  +  2  +  1)  F(l  +  1  -  1H)  and  G(2  -  Y2  -  1)  K(2  -  2  +  J$). 

154.  Lay  off  -    —  in.  (amount  specified  by  instructor)  on  any  of  the  lines  in 
Problems  143-153  from  the  first  point  toward  the  second. 

155.  Through  M(l  +1  —  3)  draw  a  line in.  long,  parallel  to  O(3  +  1 

-  2)  P(4  +  3  -  1). 

150.  Through  A (5  —2+3)  draw  a  line in.  long,  parallel  to  B(l  — 

1  +  1)C(2  -  1M  +2). 

157.  D(2  -3  +  1)  E(3M  -  3  +  2)4)  is  the  base  diagonal  of  a  square  pyra- 
mid, 2)£-in.  altitude,  in  III.     Draw  its  projections. 

(a)  Measure  one  of  its  slanting  edges. 

(&)   Measure  the  line  running  from  a  point  1%  in.  up  one  edge  to  a 

point  %  in.  up  the  adjacent  edge. 

158.  Scale   100  ft.  =  1  in.     A  rope  conveyor  is  to  be  constructed  from 
A(75  ft.,  0,  -  180  ft.)  to  a  loading  point  on  a  hill  225  ft.  high,  200  ft.  north, 

60°  east,  of  the  dumping  point.     How  much  cable  must  be  purchased 
for  a  duplex  belt,  allowing  30  ft.  for  pulleys,  slack,  etc.? 


FIG.  34. 


159.  A  pipe  runs  through  a  factory  building  from  a  point  M(5  ft.  +  10  ft. 
-  2  ft.)  to  N(30  ft.  -  2  ft.  6  in.  —  14  ft.)  in  the  floor  below. 

(a)  What  is  its  length? 

(b)  At  what  point  in  the  floor  must  the  hole  for  it  be  made  ? 

(c)  At  a  point  6  ft.  from  the  floor  hole,  measured  along  the  pipe,  is  a 
support  from  the  floor.     How  far  from  the  wall  (V)  and  the  floor  (H) 
is  this  point  of  support?     Scale  )^  in.  =  1  ft. 

160.  Fig.  34  shows  the  plan  and  elevation  of  a  plain  hip  roof.     Draw  it  to 
any  convenient  scale. 

(a)  What  angle  is  it  drawn  in?     Draw  its  P-projection. 
(6)  Locate  the  traces  of  the  roof  planes,  C  and  D. 
(c)    Wliat  is  the  length  of  the  hip  rafter  AB. 


44  PRACTICAL  DESCRIPTIVE  GEOMETRY 

(d)  Locate  a  point  on  the  roof  C,  3  ft.  below  the  ridge  and  10  ft.  to  the 
right  of  the  center. 

(e)  Locate  a  point  on  the  rafter,  3  ft.  from  F. 

161.  Scale  1  in.  =  20  ft.  A  60-ft.  stack,  3  ft.  in  diameter,  rises  from  A(60 
ft.,  0  —  35  ft.),  on  the  top  of  a  flat  roof  (H).  Five  guy-wires  are 
attached  to  the  stack  at  40  ft.  above  the  roof,  and  are  anchored  at 
points  on  the  roof,  respectively,  B(30  ft.,  0-10  ft.),  C(85  ft.,  0  - 
7  ft.  6  in.),  D(90  ft.,  0-47  ft.  6  in.),  E(70  ft.,  0-67  ft.  6  in.),  and 
F(25  ft.,  0  —  52  ft.  6  in.).  Draw  the  stack  and  guys,  and  measure  the 
true  length  of  the  guys. 


'jjH 

O                       u       y 

3,, 

^        H 

QH 

5    m« 

5    ov 

7 

8    cv 

iXb" 

Cff! 

IX< 

kH 

nv 

pH 

a 

drt 

j               1 

1                       1 

1 

!,v 

1 

y 

n    H 
In 

V 

bH 

!/ 

\ 

kv 

mH 

I 

p 

0H 

• 

av 
bv 

i 

dv 

Graphic  Layout  No.  13. — (a)   Measure  the  true  length  of  given  line  . 

(6)  Lay  off  from  one  point  of  given  line  —   —  a  point  in.  distant  on 

the  line  toward  the  second  point. 

FIFTH  FUNDAMENTAL  PROBLEM 

46.  Problem  12. — To  revolve  a  point  in  space  into  H  or  V 
about  an  axis  lying  in  that  plane. 


FIG.  35. 

Process. — Fig.  35  shows  a  point  A  in  space  and  a  line  BC  in  H. 
To  revolve  A  about  BC  into  H,  it  is  necessary  to  draw  the  radius 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     45 

of  the  circle  through  which  A  is  to  revolve.  As  will  be  seen  in 
the  picture,  Ax  is  the  radius,  and  it  is  the  hypothenuse  of  the 
right  triangle,  AaH  —  aHx.  By  laying  off  this  hypothenuse  in 
either  direction  perpendicular  to  BC  at  x,  the  revolved  position 
is  obtained. 

Analysis. — To  revolve  a  point  into  H  about  an  axis  in  H. 

1.  Draw  the  trace  of  the  plane  of  revolution  through  the  H- 
projection  of  the  point,  perpendicular  to  the  axis. 

2.  Construct  the  right  triangle  as  follows:    (1)  the  base  =  the 
perpendicular  distance  from  the  H- 

projection  to  the  line;  (2)  the  alti- 
tude =  the  F-projection  to  GL. 

3.  Lay  off  the  hypothenuse  of  this 
triangle  from  the  axis  on  the  trace 
of  the  plane  of  revolution. 

Construction. — Let  A  (Fig.  36)  be 
the  given  point  and  BC  the  given 
line  in  H. 

1.  Through  aH  draw  xaiH  perpen- 
dicular to  bHcH. 

2.  Construct  the  triangle  aHx  — 
aHz(  =  avy). 

3.  Lay  off  xz  on  xaiH,  locating  aiH. 
Conclusion. — aiH  is  the  revolved 

position  of  A  about  BC. 

Notes. — To  revolve  a  point  into  V  about  an  axis  in  V,  make  an  interchange 
of  H  and  V  in  the  foregoing. 

2.  If  the  point  is  directly  over  the  axis,  the  base  of  the  triangle  becomes 
zero,  and  the  hypothenuse  will  be  equal  to  the  altitude. 

3.  The  "trace  of  the  plane  of  revolution"  and  "the  projection  of  the  path 
of  revolution"  will  be  abbreviated  hereafter  to  "the  path  of  revolution." 


47. 


EXERCISES 


162.  Revolve  A (2  -  1  +  3)  and  B(4  +  2  -  2^)  into  H  about  C(l,  0-2) 
D(5,  0  -  H). 

163.  Revolve  E(l  +  1  -  2)  and  F(4  -  1  +  2)  into  V  about  G(2  +  3,  0) 
K(5  +  M,  0). 

164.  Revolve  M(l  -  2  -  1),  N(2  +  1  -  2),  O(3  +  1  +  1),  and  P(4  -  1  + 
1H)  about  GL,  (1)  into  7;  (2)  into  H. 

165.  Revolve  K(2  +  1^-1)  about  L(l  +  3,  0)  N(5,  0,  0)  into  V.  Revolve 
M)3  -2  +  1)  about  P(2,  0,  0)  O(5,  0-1)  into  H. 

166.  Locate  A(l  +  1,  x)  and  B(2,  y  -  1)  in  T(l  +  3)  5(1  -  2),  and  revolve 
into  V  aboutVT,  and  into  H  about  HT. 


46 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


167.  Locate  C(4  +  2,  x)  and  D(2,  y  -  1)  in  8(5  +  3)  1(5  -  1),  and  revolve 
intoF  about  VS,  and  into  H  about  HS. 

168.  Locate  E(2,  x  -  2)  and  F(4  +  1,  y)  in  U(  +  2)  oo(  -  1),  and  revolve 
them  about  VU  and  HU  into  V  and  //. 


i     !,v 


k"  i 


1      g 

1 

ifv 


Graphic    Layout  No.  14. — Revolve  the  points  in  Ex. 
given  line  into  the  plane  of  the  latter. 


about  the 


169.  Locate  K(2  -  2,  x)  and  L(4,  y  +  1)  in  W  (-  1)  «(  +  3),  and  revolve 
them  about  VW  and  HW  into  7  and  H. 


1            s 

2           ,0^ 

3    ,cy 

|    VW 

4 

5-^"^" 

A>''      i^ 

~£o  *   ic. 

H 

XX 

'k 

S        ' 

2£r*^_ 

IP" 

l/^ 

^v^ 

\.v  | 

1 

•  K     ' 

1 

1 

HW  

> 

^^^^^  *  ^^^^^^ 

if      \/V 

Graphic  Layout  No.  15. — Locate  the  points  in  plane  —  —  and  revolve  them 
into about  the trace. 

170.  Locate  O(  +  2)  and  P(  +  1)  in  the  F-trace  of  T(l  +  3)  5(1  -  2)  and 
•  revolve  them  into  H  about  the  H  -trace. 


HT 

b 


VT 


Graphic  Layout  No.  16. — Locate  the  plane  figure in  its  plane,  and  re- 
volve about  one  of  the  traces  to  show  its  true  size. 

171.  Locate  A(2-l,  z)  andB  (4,  0    +   2)  in  S(  -  1)  °°  (+  2),  and  revolve 
A  into  H  about  HS,  and  B  into  V  about  VS. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES    47 


172.  Revolve  the  triangle  M(1J^  +  1^-1)  N(2^  +  1  - 

1/4  —  %}  into  V  about  the  F-trace  of  its  plane.     Will  this  give  its 
true  size,  and  the  true  angles  at  the  corners? 

173.  Revolve  the  parallelogram   A  (3  -1  +  1)  B(3  %  -%  +  2)    C(5  - 
2M  +  1M)  D(4^  -  2^  +  M)  into  #  about  the  //-trace  of  its  plane. 

Will  this  give  its  true  size? 

MEASUREMENT  OP  ANGLES 

48.  Problem  13.  —  To  measure  the  true  angle  between  any 
two  intersecting  lines.  . 


FIG.  37. 


Two  intersecting  lines  lie  in  one  plane,  and,  if  that  plane  be 
folded  so  as  to  coincide  with  H  or  V,  the  revolved  projections 
will  show  the  true  relation  of  the  lines. 

Analysis. — 1.  Pass  a  plane  through  the  two  lines. 

Note. — One  trace  only  need  be  obtained. 

2.  Revolve  the  plane  about  its  .H-trace  into  H,  or  about  its 
Y-trace  into  V. 

Note. — Revolving  the  various  points  in  a  plane  is  equivalent  to  revolving 
the  plane. 

Conclusion. — The  angle  between  the  lines  in  revolved  position 
is  the  measure  of  the  actual  angle. 

Construction. — Let  AB  and  BC  (Fig.  37)  be  the  two  inter- 
secting lines  whose  angle  is  to  be  measured. 


48 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


1.  Pass  the  plane  through  the  two  lines;  HT  is  the  necessary 
trace. 

2.  Revolve  AB  and  BC  about  the  trace. 
Conclusion.  —  The  required  angle  is 


Note  1.  —  It  is  usual  to  consider  the  acute  angle  as  the  measure,  although 
the  obtuse  angle  is  not  incorrect. 

Note  2.  —  In  case  any  of  the  necessary  piercing  points  are  inaccessible,  it  is 
proper  to  draw  as  many  auxiliary  lines  from  one  line  to  the  other  as  may  be 
needed  to  obtain  the  trace. 

SPECIAL  CASES 

49.  Case  I.  —  When  one  side  is  parallel  to  H. 

Analysis.  —  1.  Revolve  the  oblique  line  about  the  horizontal 
line  as  an  axis  until  it  becomes  parallel  to  H. 


FIG.  37a. 


Conclusion. — The  projected  angle  is  now  equal  to  the  true  an- 
gle, since  both  are  parallel  to  H,  and  their  relations  are  unchanged. 

Construction. — Required  to  measure  the  angle  ABC  (Fig. 
37o),  AB  being  parallel  to  H. 

1 .  Revolve  C  about  AB  as  an  axis  into  the  horizontal  plane  of  AB . 

Note  1. — This  can  be  easiest  done  by  transferring  GL  to  avbv,  and  ap- 
plying Problem  12. 

Note  2. — The  same  process  can  be  performed  when  one  of  the  lines  is 
parallel  to  V  or  P. 

Note  3. — If  preferred,  the  regular  processes  can  be  used  instead  of  this 
special  one. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     49 

50.  Case  II. — When  one  side  is  perpendicular  to  H,  V,  or  P. 
Construction. — Project  the  lines  on  Q,  parallel  to  the  two  lines. 

The  Q-projection  of  the  angle  is  the  true  size  of  the  angle.     See 
Fig.  376. 

51.  Case  III. — When  both  sides  are  parallel  to  P. 
Conclusion. — The  P-projection  of  the  lines  will  show  the  angle 

in  its  true  size. 

52.  Problem  14. — To  find  the  projections  of  the  bisector  of 
any  angle. 

Construction. — Let  ABC  (Fig.  38)  be  the  given  angle.     To 
find  the  bisector: 


FIG.  38. 

1.  Obtain  HT  (or  VT,  if  more  convenient). 

2.  Revolve  ABC  into  H  about  HT,  making  the  true  angle 


3.  Bisect  aHbiCi.     The  bisector  is  dibi. 

4.  Draw  an  auxiliary  line  aHCi,  intersecting  bidi  in  di. 

5.  Revolve  aHc,  back  to  its  original  position,  aHcH,  avcv. 

6.  Revolve  d,  back  to  dH  on  aHcH,  and  project  to  dv  on  avcv. 

7.  Connect  the  projections  of  B  and  D. 
Conclusion.  —  BD  is  the  bisector  in  original  position. 

4 


50 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Note. — If  the  angle  can  be  measured  by  any  of  the  special  cases,  the  bi- 
sector may  be  found  by  the  same  process. 


53. 


EXERCISES 


I         i 


l«     I 


B 

V 


I     I 

I I 


leH  ! 


\lfH 


Graphic  Layout  No.  17. — (a)  Measure  angles  between  given  lines  in  Ex. 
— .     (b)  Draw  the  projections  of  the  bisector  of  the  angle  in  Ex. . 


Measure  the  angles  included  between  the  following  lines: 

174.  A(2  +  2  -  2)  B(3M  +  1  -  M)  and  BC(4^  +  3  - 

175.  D(l  -  2  +  3)  E(4  -  iy2+  \Yti  and  EF(2  -1  +  1). 

176.  G(l  +  2>£  -  1)  K(3>£  +  1M  -  2)  and  KL(2  +  1  - 

177.  M(1H  +  3  -  2)  O(2  +  1  -  M)  and  OP(5  +  Y2  - 

178.  A(2  +  3-  2)  B(2  +  1  -  1)  and  BC(3H  +2-3). 

179.  D(4  +  1  -  1)  E(4  +  2H  ~  2)  and  EF(3  +  1  -  1). 

180.  G(3  +  1  -  3)  K(3  +  3  -  1)  and  KL(2  +  3-1). 

181.  M(l  +  2  -  1)  N(l  +  2  -  3)  and  NO(2^  +  M  ~  D- 

182.  A(l  +  1H  -  2)  B(4  +  1H  -  2)  and  BC(2  +  1-1). 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     51 


183.  D(3  +  3  -  Ifa)  E(l  +  1  -  K)  and  EF(3  +  1  -  3)  by  revolving  into 
the  profile. 

184.  G(2  +  IY2  -  3)  K(4  +  U2  -  1)  and  KL(3  +  3  -  1^). 

185.  M(l  -  2^+  1).N(4  -  y2  +  1)  and  NO(2  -1+2). 

186.  A(2  +  2  +  2)  B(4  +  K  +  K)  and  BC(3  -  1  -  1). 

187.  D(4  -  2-  2)  E(2  +  y2  +  M)  and  EF(3  +2-  1). 

188.  G(3  +  1  +  1)  K(3  -  1^  -  iy2}  and  KL(1  -  1  -  1) 

189.  M(4,  0,  0)  N(2  -  1  +  1J£)  with  GL. 

Measure  the  angle  between  the  traces  of  the  following  planes: 

v 

190.  T(l  +  1)  4(1  -  2M). 
>191.  S(l  +3)4(1+2). 

192.  R(4  -  1^)  1(4  +  2). 

193.  Q(l  +  2)  3(5  -  2). 

194.  U(4  +2)  3(2-  2). 

195.  W(5  +  3)  3K(1  -  1). 

196.  T(l  +  2)  1(3  -  3)  and  S(2  +  1^)  5(5  -  3) 

197.  R(l  +  3)  (5  +  1K)(1  -  2)  (5  -  1). 

198.  W(l  +  3)(5  +  iy2)(l  +  2)(5  +  1). 


Graphic  Layout  No.  18. — (a)  Measure  the  angle  between  the  H-  and  V- 

traces  of  plane .     (ft)  Draw  the  bisector  of  the  angle  beween  the  traces 

of  plane . 


Measure  the  angles  in  the  following  triangles: 


A  (2  -  1  +  3)  B(3  -  2  + 
D(2^  +  1  -  1)  E(4  +  3  - 
G(2  +  2  -  2)  K(2  +  1  - 
M(2^  -1Y2.+  2)  0(3  + 
A(2+  2  +  2)  B(2M  +  1 
D(5  -  1  -  1)  E(4  -  2-  2) 


C(5-  1%  +  2). 

F(4  +  2  -  2). 
L(3  +  2  -  2). 
-  1)  P(4  -  %  +  1). 
C(4  +  1>£ 
+  1  - 


199. 
200. 
201. 
202. 
203. 
204. 

205.  G(2  +  1^,  0)  K(3M,  0,  0)  L(3,  0  -  2). 

206.  Measure  the  angles  and  true  size  of  the  parallelogram  M(3  +  K  — 

iH)  N(3  +  i  -  y2)  0(4  +  2  -  i)  P(4  +  iy2  -  2). 

207.  Measure  the  angles  of  the  pentagon  A(3^  +  Vi,  x)  B(4  +  !M,  y) 
C(4>2X  +  M,  z)  D(5>i  +  %,  m)  E(5  +  1H,  n)  that  lies  in  the  plane 
T(6+3)l(5  - 


52  PRACTICAL  DESCRIPTIVE  GEOMETRY 

208.  Prove  graphically  that  A  (2  +  %  -  1)  B(3  +  2  -  2)  C(4^  +  2  -  ^) 
is  a  right  angle.     Write  an  explanation. 

209.  Draw  a  line  from  D(2  +2  -  1)  intersecting  E(2  +1  -  2)  F(5  +  2  - 
K)  at °(30,  45,  60,  75). 

210.  Draw  a  line  from  G(3  +  1-1)  intersecting  K(l  +  2  -  3)  L(4  +  2  - 
3)  at °(30,  45,  60,  75). 

211.  Draw  a  line  from  M(2  +  ^  -  2)  intersecting  N(l  -f  2  -  1)  O(3  + 

1  -  1)  at °(30,  45,  60,  75). 

212.  Draw  a  line   from  P(2  +  2  -  1)  intersecting  A(3  +  1-2)  B(3  + 

2  -  K)  at °(30,  45,  60,  75). 

213.  Draw    a    perpendicular    from    C(2  +  2  -  K)    to    D(2  +  %  -  ^) 
E(2  +  2  -  1^). 

214.  Measure  the  angle  between  two  adjacent  edges  and  the  angle  between 
an  edge  and  a  base  line  of  the  pyramid  given  in  Ex.  157,  Art.  45. 

215.  Measure  the  angle  that  E(2  +  ^  -  2^)  F(4  +  2  -  1)  makes   (1) 
with  its  own  F-projection,  (2)  with  its  //-projection. 

216.  Draw  the  roof  in  Fig.  34  to  a  definite  scale,  and  measure  the  following 
angles : 

(a)  Hip  rafter  EF  with  hip  rafter  EG. 
(6)   Hip  rafter  EF  with  eave  FG. 

(c)  Hip  rafter  AB  with  eave  AG. 

(d)  Hip  rafter  AB  with  ridge  BE. 

217.  Draw  the  roof  in  Fig.  34  to  a  definite  scale,  and  measure  the  true  size 
of  the  roof  planes  C  and  D. 

Compute  their  areas  from  the  scale. 

What  would  it  cost  to  roof  it  at  $1.25  per  square  yard? 

218.  The  F-trace  of  T(l  +  3)  4(x,  y)  makes  60°  with  its  tf -trace.     Draw 
the  H-trace. 

219.  The  ff -trace  of  S(x,  y)  3(5  -  1>£)  makes  120°  with  its  F-trace.     Draw 
the  F-trace. 

220.  A  ray  of  light  passes  through  A(l  +  1  —  2)  and  is  reflected  from  H 
to  B(4  +  lJ-£  —  1).     At  what  point  and  angle  does  it  strike  HI 

221.  A  ray  of  light  passes  through  C(l  +2  —  1)  and,  after  reflecting  from 
V  and  H,  passes  through  D(5  +  ^  —  2).     At  what  points  and  angles 
does  it  strike  H  and  F? 

222.  Draw  a  line  from  E(4,  x  -  1)  in  T(5  +  3)  2(5  -  2)  making  60°  with 
the  ff-trace. 

223.  Draw  a  line  from  K(2  +  1  -  1^)  in  the  triangle  KL(2  +2-2) 
N(3  +2  —  2)  making  60°  with  the  opposite  side. 

224.  Find    the    center    of    the    triangle    A(2  +  1M  -  1)  B(3  +  l  -  Y2) 
C(3K  +3-2). 

225.  Find   the  center  of   the  triangle  D(l^  -2  +  1)  E(2^  -  1  +  H) 
F(3K  -  1M  +  2). 

Draw  the  projections  of  the  bisectors  of  the  following  angles  (Ex.  226- 
247): 

226.  A (4  +  2M  -  1)  B(2  +  1  -  iy2)  C(4  +  1%  -  K). 

227.  D(2  -  1+  3)  E(2M  -  3  +  \Y^)  F(4  -  IK  +  D- 

228.  G(2  +  2  -  1)  K(3^  +  3  -  3)  L(2  +  M 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     53 


229.  M(4  -  H  +  2)  N(2  -  3  +  2K)  O(4  -IK  +  K). 

230.  A  (2  +  IK  -  3)  B(2  +  K  -  %)  C(2  +  3  -  2). 

231.  D(2  -  1  +  K)  E(2  -  Y2  +  \Y2)  F(2  -  3  +  y2) 

232.  G(l  +  K  -  3)  K(2  +  1  -  1)  L(4  +  2-  H). 

233.  M(l  +  2^  -  2)  N(l  +  1  -  1)  O(3  +  2  -  2^). 

234.  A(2K  +  2^  -  K)  B(2^  +  H  -  1%)  0(1  +  2  -  1). 

235.  D(2+  1^  -  1)  E(5,  0,  0)  F(2,  0,  0). 


t> 


'I     ! 


8 


Graphic  Layout  No.  19.  —  (a)  Measure  the  angle  of  the  polygon 
(6)  Locate  the  center  of  triangle  —   —  (1,  2,  3,  4,  8,  9). 


236.  G(l  -  1+  K)  K(l  -  2y2  +  2)  L(l  -  1^ 

237.  Between  the  H-  and  F-traces  of  T(l  +  3)  5(1  - 

238.  Between  the  H-  and  F-traces  of  S(l  +  3)  2(5  -  1). 

239.  Between  the  H-  and  F-traces  of  R(5  +  3)  1(5  +  2). 

240.  Between  the  F-  and  P-traces  of  R(5  +  3)  1(5  +  2) 

241.  Between  the  F-  and  P-traces  of  Q(+  1)  «,  (-  2). 


M 


d>f 


5av 


m 


VT 


ik" 


ixH 
1=GL 


Graphic  Layout  No.  20. — (a)  Draw  a  perpendicular  from  point to 

line  —  —  in  Ex. .  In  Ex.  6-8  locate  the  point  in  the  plane,  and  draw 

the  perpendicular  to  the trace.  (6)  From  the  given  point  in  Ex. 

—  draw  a  line  making °  (15,  30,  45,  60,  75)  with  the  given  line  or 

trace. 


242.  Between  the  H-  and  F-traces  of  U(5  -f-  3)(1  -  1)(5  + 

243.  Between  the  H-  and  P-traces  of  U(Ex.  242). 

244.  Between  the  H-  and  F-traces  of  W (2  +  3)  3(4  -  3). 

245.  Between  the  H-  and  P-traces  of  W(2  +  3)  3(4  -  3). 

246.  M(l  +  2  -  3)  N(3  +  1  -  K)  and  its  F-projection. 

247.  O(4  -  1  +  3)  P(2  -  3+  1^)  and  its  ^-projection. 


54 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


64.  Problem  16. — To  find  the  shortest  distance  from  a  point 
to  a  line. 

TWO  METHODS 

First  Method. — Analysis. — 1.  Pass  a  plane  through  the  given 
point  and  line. 

2.  Revolve  the  point  and  line  about  the  trace  of  the  auxiliary 
plane,  as  in  the  preceding  problems. 


FIG.  39. 

3.  Erect  the  perpendicular  from  the  point  to  the  line. 

Conclusion. — This  perpendicular  is  the  true  shortest  distance, 
as  the  point  and  line  are  in  their  true  relation. 

Construction. — Let  C  (Fig.  39)  be  the  given  point  and  AB  the 
given  line. 

1.  Pass  the  plane  T  through  AB  and  C,  obtaining  HT  only. 

2.  Revolve  AB  and  C  about  HT  to  a,  b,  and  c. 

3.  Erect  the  perpendicular  Cidi  from  Ci  to  aibi.     This  is  the 
shortest  distance. 

Note. — If  the  shortest  line  is  desired,  Cidi  may  be  revolved  back  to  the 
original  projections  of  the  line  as  shown  in  the  figure. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     55 

66.  Second  Method  by  auxiliary  planes. 

Let  AB  (Fig.  40)  be  the  given  line  and  C  the  given  point. 

Construction. — 1.  Project  AB  and  C  on  Q,  which  is  taken 
parallel  to  AB  (and  perpendicular  to  H,  of  course). 

2.  Draw  QX  perpendicular  to  aQbQ,  and  project  AB  and  C  on  Q. 

Conclusion. — The  distance  from  cx  to  axbx  (a  point  in  projec- 
tion) is  the  true  distance.  Projecting  back,  CD  is  the  shortest 
line. 


FIG.  40. 


56. 


EXERCISES 


Find  the  shortest  distance  from  the  following  points  to  the  following 
lines : 

248.  A(4+  3  -  1)  to  B(2  +  2^  -  Y2)  C(4  +  1  -  3). 

249.  E(l  -  3  +  2H)  to  D(1K  -  M  +  1)  F(±H  -3+2) 

250.  G(3  +  2  -  3  to  K(l  -  2  +  3)  L(4  -  1+  1). 

251.  M(3  +  1  -  3)  to  N(2  +  2,  0)  O(4  +  1,  0).' 

252.  P(3+2  -  \Y2}  toGL. 

253.  A(2  +  1  +  1)  to  B(l  -  3  -  3)  C(3  -  1  -  1). 

254.  D(3  -  1  -  1)  to  E(2  +  2  +  2)  F(4  +  Y2  +  Y^. 


56 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


255.  G(l  +  3,  0)    on   the    F-trace   of   T(l  +  3)  4(1  +  2)    to    HT.     Also 
K(2K,  0  -  1)  on  HT  to  VT. 

256.  A  2-in.  steam  pipe  runs  through  a  basement  from  a  point  in  the  floor, 
M(10  ft.,  0,  -  18  ft.  6  in.)  to  a  point  on  one  wall,  N(50  ft.  +  13  ft., 
0).     At  a  point  O(22  ft.  +  18  ft.  -  2  ft.)  in  the  ceiling  it  is  desired  to 
make  the  shortest  possible  connection,  using  1^-in.  pipe.     Where  will 
the  joint  be  made,  and  how  long  will  the  branch  pipe  be,  allowing  3  in. 
for  connection? 


1  av      ,v 

2          eH 

3fv 

4 

5av 

6 

7 

8 

NT] 

rr 

K 

*] 

) 

mv 

' 
nv 

V 

m 

bV  ! 

K> 

VT 

/" 

\ 

Jb1 

1        IH 

If"     ! 

mH 

nH 

b"  1  " 

^\ 

Vx 

!/ICH 

\ 

V 

1          k" 

9H 

XH 

a" 

HT 

^^.e 

MN=GL 

Graphic  Layout  No.  21. — (o)  Measure  the  distance  from  the  given  point 

in  Ex.  to  the  given  line.     (In  Ex.  6-8  locate  the  point  in  the  plane 

given,  and  measure  the  distance  to  the (H-  or  F-trace).     (6)  Draw  the 

projections  of  the  shortest  possible  line. 

FIGURES  IN  OBLIQUE  PLANES 

57.  Problem  16. — To  draw  the  projections  of  any  plane 
figure  in  any  oblique  plane. 

METHODS 

There  are  two  important  processes  of  achieving  this  result. 
The  student,  by  mastering  both,  will  have  complete  command 
over  any  situation  that  will  arise.  The  difference  in  the  method 
is  mainly  in  the  manner  of  counter-revolving  the  points. 

Analysis. — Assuming  that  a  given  point  in  a  given  oblique  plane 
is  to  be  the  center  of  some  regular  polygon  lying  in  the  plane. 

1.  Revolve  the  given  point  about  the  trace  of  the  given  plane, 
as  in  Problem  12. 

2.  Construct  the  required  polygon  in  its  true  size  about  the 
revolved  position  of  the  center. 

3.  Counter-revolve  the  points  (corners)  of  the  polygon  into  the 
original  position  of  the  plane. 

Note. — The  above  is  the  analysis  for  both  methods. 


58. 


METHOD  BY  ORDINARY  REVOLUTION 


Construction. — Let  O  (Fig.  41)  be  a  point  in  T.  Let  it  be 
required  to  draw  the  projections  of  a  1-in.  square  in  T,  about  0 
as  a  center,  having  two  of  its  sides  parallel  to  H. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     57 

1.  Revolve  O  about  HT  to  OL 

2.  Construct  the  square  aibiCidi  in  its  true  size  about  o,  as 
a  center  with  two  sides  parallel  to  HT. 

3.  Draw  through  01  a  line  through  any  point  or  points  of  the 
square  (b1o1c1x),  extending  it  to  HT. 

4.  Counter-revolve   this   line  to  the  position  XOH.     (This  is 
possible,  for  X,  being  in  the  axis  will  remain  stationary,  and  QI 
will  counter-revolve  to  OHOV.) 

5.  Counter-revolve  d  and  bi  to  this  line  on  paths  perpendicular 
to  HT,  also  &i  and  di  by  the  same  method. 


FIG.  41. — A  square  drawn  in  an  oblique  plane. 

6.  Locate  the  F-projections  of  A,  B,  C,  and  D  by  any  of  the 
known  methods. 

The  square  will  then  be  drawn  according  to  its  specifications. 


59. 


AUXILIARY  PLANE  METHOD 


Construction. — Let  O  (Fig.  42)  in  T  be  the  given  center  of  a 
1-in.  regular  hexagon. 

(Note. — When  the  diameter  of  a  hexagon  is  given,  it  means  the  diameter 
of  the  circumscribed  circle,  long  diameter,  unless  short  diameter  or  diameter 
"across  the  flats"  is  specified.) 

1.  Assume  Q  perpendicular  to  HT  at  any  convenient  point. 


58 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  Obtain  the  Q-trace  of  T  and  the  Q-projection  of  0. 

3.  Revolve  O  about  HT  to  QI  in  the  manner  shown  in  Fig.  42. 

4.  Construct  the  required  hexagon  about  QI  as  a  center. 

5.  Counter-revolve  the  corners  of  the  hexagon:     (1)  to  QT; 
(2)  to  the  //-projection;  (3)  to  the  F-projection. 

Note  1. — The  arrows,  in  Fig.  42,  show  the  direction  of  this  operation  in 
the  case  of  point  A. 

Note  2. — The  F-projections  of  the  various  points  are  the  same  distance 
from  GL  as  the  Q-projections  are  from  HQ. 


FIG.  42.  —  A  hexagon  drawn  in  an  oblique  plane  by  the 

method. 


auxilliary  plane" 


60.  Supplementary  Problem.  —  To  draw  the  projections  of  a 
circle  in  an  oblique  plane. 

Construction.  —  Let  0  (Fig.  43)  be  the  center  of  a  1^-in.  circle 
lying  in  T. 

1.  Revolve  0  about  HT  to  Oi. 

2.  Draw  the  circle  full  size  about  01  as  a  center. 

3.  Draw  the  diameters  parallel  and  perpendicular  to  HT,  aibi 
and  Cidi  respectively. 

4.  Counter-revolve  the  points  ai,  bi,  ci,  and  di.     They  will 
be  the  four  vertices  of  the  ellipse,  which  is  the  //-projection  of 
the  circle.     The  major  axis  will  be  aHbH  and  the  minor  axis  cHdH. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     59 

The  ellipse  may  be  drawn  on  these  axes  in  any  desired  manner, 
preferably  the  "trammel  method."     (See  Art.  147.) 

For  the  F-projection,  revolve  O  about  VT  and  counter-revolve 
the  points  ej,  fi,  gi,  and  kx  in  the  same  way;  evfv  will  then  be  the 
major  axis,  and  gvkv  will  be  the  minor  axis. 


FIG.  43. — The  projections  of  a  circle. 


61. 


EXERCISES 


Draw  the  projections  of  the  following  plane  figures  in  the  planes  given 
according  to  the  stated  conditions: 

257.  Equilateral  triangle,  l^-in.  sides.     T(2  +  3)  2(5  -  1^)-     Center  at 
A(3K  +  1,  x)  in  T. 

258.  Triangle  Ex.  257.     U(2  +  2)  4(4  -  3).     One  side  ||  to  H,  1  in.  from 
HU. 

259.  Triangle  Ex.  257.     S(5  +  2)  2(5  -  1H)-     Center  at  O(4  -f  1,  x)  in  S. 

260.  Triangle  Ex.  257.    T(2  -  3)  4(2  +  2).     One  side  ||  to  V,  1  in.  from  VT. 

261.  Triangle  Ex.  257.     U(+  2)  «  (-  1).     Center  at  M(3,  y  -  K)- 

262.  Triangle  Ex.  257.     W(  -  3)  » (+  2).     One  side  ||  to  H,  1  in.  from  VW. 


60 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


263.  Triangle  Ex.  257.  X(+  J$)  °°  C~  2).     Center  at  N(3  +  1^,  y). 

264.  Triangle  Ex.  257.  Q(l  +  3)  3(5  -  1).     Center  at  K(4  +  1,  x). 

265.  Triangle  Ex.  257.  R(l  +  1)   (5  +  2)  (1  -  %)  (5  -  1).     Center    at 
F(3  +  1,  x). 

266.  Triangle  Ex.  257.  8(5  +  2)  1(5  +  1).     Center  at  D(3,  x  -  1). 


$\!l 

L-//J 
(i) 

V?CTfn  9  O  Q  Q 

<..//>!     k-//M        ivp         K/p       k//M      U//'»l      U//'J 

(2)          (3)          (4)            (5)         (6)          (7)          (8) 

PLANE  FIGURES 

37 
t 

^ 

/ 
1 

4 
~HW~ 

5 

"~VX 
HX 

V 

\ 

A- 

A 

~5? 
S§> 

"~VW~ 

/ 

7 

8^ 
^^^ 

9 

O" 

> 

% 

11          / 

^ 

^'^ 

12  ^^ 

r^s 

vb" 

-_vz  

HZ 

-^J^ 

^\ 

J 
1 

frrflfjlltr.    T.nvrmt    TTr>      9.9.          TlrQwr    t.Vif>    r»rr»ip>r»t.ir>na    r>f   nlano   firniro                   in 

given  plane  ,  under  one  of  the  following  conditions,  as  selected  by  the 

instructor: 

(a)  Assume  a  point  in  the  given  plane,  as  the  center  of  the  given  plane 
figure. 

(6)  Assume  a  line  in  the  given  plane  as  one  of  the  sides  of  the  plane 
figure. 

(c)  Assume  a  line  in  the  given  plane  as  the  diagonal  or  diameter  of  the 
given  plane  figure. 

(d)  Draw  one  corner  of  the  given  figure  in  H,  and  one  side  inclined ° 

(10,  15,  30,  40,  45)  to  the  tf-trace. 

Note. — V  may  be  substituted  for  H  in  condition  (d). 

(e)  Circumscribe  a  circle  about  the  plane  figure in  plane . 

(f)  Inscribe  a  circle  in  the  plane  figure in  plane . 

267.  Triangle  Ex.  257.     Y(l  +  2^)  3(5  -  2}$).     Center  at  A(2H,  x  -  1). 

268.  Triangle  Ex.  257.     P (4  +  3)  4(4  -  3).     One  side  ||  to  H,  1  in.  below 
H  in  ///. 

269.  Substitute  a  right  triangle  1  in.X  1M  in.  in  any  of  the  planes  in  Ex. 
257-268,  under  the  same  conditions. 

270.  Inscribe  a  circle  in  any  of  the  triangles  in  Ex.  257-269. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     61 

271.  Substitute  one  of  the  following  polygons  for  the  triangle  in  Ex.  257-268: 
(a)  \Y±  in.  square. 

(6)    1-in.  X  1^2-in.  rectangle. 

(c)  1^-in.  regular  pentagon. 

(d)  1^-in.  regular  hexagon. 

(e)  1%-in.  circle. 
(/)    IJ^-in.  octagon. 

272.  Draw  the  projections  of  polygon in  Ex.  271  in  any  of  the  planes 

given  in  Ex.  257-268  with  one  corner  in  H  and  another  corner  ^  in. 
from  the  fl-trace. 

273.  Triangle  in  Ex.  269  in  any  of  the  planes  in  Ex.  257-268,  with  its  hy- 
pothenuse  ||  to  V,  1  in.  from  the  F-trace. 

274.  Square,  Ex.  271  (a)  in  any  of  the  planes  in  Ex.  257-268,  with  its  diag- 
onal ||  to  H,  Y±  in.  above  or  below  H. 

275.  Hexagon,   \Y±  in.  short  diameter,  with  a  M-in.  circle  in  the  center, 
drawn  in  any  of  the  planes  in  Ex.  257—268,  with  its  center  at  any  chosen 
point  in  the  given  plane. 

276.  Circumscribe  a  circle  about  any  of  the  foregoing  polygons. 

277.  Square,  Ex.  271  (a)  in  any  of  the  planes  in  Ex.  257-268,  with  one  corner 
in  V,  and  a  side  inclined  30°  to  the  F-trace. 

278.  The  line  M(3  +  1,  x)  N(4  +  M,  y)  in  R(l  +1)   (5  +  2)  (1  -  M) 
(5  —  1)  is  one  of  the  diameters  of  a  circle  in  R.    Draw  the  projections 
of  the  circle. 

279.  An  ellipse,  major  axis  A (2,  0-1)  B(3K,  0  -  1),  minor  axis  C(2%, 
0  ~  M)  D(2%,  0  —  lf£)  is  the  projection  of  a  circle  lying  in  a  plane, 
whose  #-trace  is  E(l,  0-2)  F(4,  0  -  2).     Draw  the  F-trace  of  the 
plane  and  the  F-projection  of  the  circle. 

280.  A  1^-in.  circle  whose  center  in  M  (3,  0  —  1)  is  the  fl-projection  of  an 
ellipse  lying  in  any  of  the  planes  in  Ex.  258,  259,  261,  263,  264,  265, 
266,  267.     Draw  the  F-projection  and  the  true  size  of  the  ellipse. 

281.  The  roof  of  a  tower  is  a  hexagonal  pyramid,  base  14  ft.,  and  altitude 
18  ft.     In  each  of  the  three  front  faces  are  rectangular  windows,  2  ft. 
9  in.  X  6  ft.  3  in.,  each  3  ft.  3  in.  from  the  tower  base.     Draw  the 
projections.     Scale  %  in.  =1  ft. 

SIXTH  FUNDAMENTAL  PROBLEM 

62.  Problem.  17. — To  find  the  line  of  intersection  of  two  planes. 
Axioms. — 1.  'The  intersection  of  two  planes  is  a  straight  line. 
2.  Two  points  determine  a  straight  line. 

Analysis. — 1.  Find  both  projections  of  the  point  where  the 
F-traces  of  the  given  planes  intersect. 

2.  Find  both  projections  of  the  point  where  the  H  -traces 
intersect. 

3.  Join  the  respective  projections  of  these  two  points,  which 
are  common  to  both  planes. 

Conclusion. — The  line  thus  drawn  is  common  to  both  planes, 
and  therefore  their  intersection. 


62 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Note. — The  H-projection  of  any  F-trace  is  in  GL,  and  therefore  the  ^-pro- 
jection of  any  point  in  the  F-trace  is  in  GL. 

Proof. — The  F-traces,  being  each  a  line  in  one  of  the  given 
planes,  intersect  in  a  point  common  to  both  planes,  and  therefore 
in  the  line  of  intersection.  Similarly  for  the  intersection  of  the 
^-traces.  Having  found  these  two  points,  the  line  of  intersec- 
tion is  determined. 

Construction. — Left  to  the  student.     Refer  to  Fig.  44. 


FIG.  44. 


63.  Special  Cases. — When  the  traces  do  not  intersect  within 
the  limits  of  the  problem,  or  not  at  all. 

Analysis. — 1.  If  one  pair  of  traces  intersect,  locate  their  point 
of  intersection. 

2.  Draw  an  auxiliary  plane  that  will  intersect  both  given  planes. 
The  lines  of  intersection  of  the  auxiliary  and  given  planes  will 
intersect  in  a  point  on  the  required  line  of  intersection. 

3.  If  neither  pair  of  traces  intersect,  draw  a  second  auxiliary 
plane  which  will  determine  another  point  on  the  required  line, 
thus  determining  the  line  of  intersection. 

Construction. — For  most  oblique  planes,  such  as  the  two  shown 
in  Fig.  45,  it  is  quickest  and  most  convenient  to  use  planes 
parallel  to  H  or  V  as  auxiliaries.  Such  planes  have  one  trace 
only,  and  that  is  parallel  to  GL.  Such  planes  intersect  oblique 
planes  in  lines  parallel  to  their  traces;  that  is,  a  plane  parallel 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     63 

to  V  will  intersect  an  oblique  plane  in  a  line  parallel  to  its  V- 
trace,  etc.     Let  the  student  give  the  reason  from  Solid  Geometry. 

1.  Draw  VQ  (Q  ||  to  #).     Q  will  cut  CM  and  DM  from  T  and 
S  respectively. 

2.  Locate  mv  and  mH,  the  intersection  of  CM  and  DM. 

3.  Draw  HR  (R  ||  to  7).     R  will  cut  AO  and  BO  from  T  and 
S  respectively. 


4.  Locate  ov  and  OH,  the  intersection  of  AO  and  BO. 

5.  Draw  ovmy  and  oHmH,  the  projections  of  the  required  inter- 
section. 

64.  Case  II. — When  it  is  convenient  to  use -an  auxiliary  plane 
parallel  to  one  of  the  given  planes. 

Fig.  46  shows  a  plane  Q,  parallel  to  S,  being  used  to  determine 
the  direction  of  the  line  of  intersection  of  T  and  S. 

Construction. — 1.  Pass  a  plane  Q  parallel  to  S,  so  that  its  inter- 
section XY  with  T  will  come  inside  the  problem  limits. 

2.  Draw  XY,  the  intersection  of  T  and  Q. 


64 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


3.  Find  M  (one  point  on  the  intersection  of  T  and  S). 

4.  Draw  MN  parallel  to  XY. 

Conclusion.  —  MN  is  the  required  intersection  of  T  and  S. 


FIG.  46. 

65.  Case  III. — When  both  planes  are  parallel  to  GL. 

Although  the  planes  intersect,  then-  traces  are  all  parallel. 
Inasmuch  as  both  planes  are  parallel  to  GL,  their  intersection 

VT I 


VS 


HT 

HS 


FIG.  47. 


will  also  be  parallel  to  GL.     (See  Solid  Geometry  problems). 
Hence  only  one  point  on  the  intersection  is  required. 

Figs.  47  and  48  show  two  different  auxiliary  planes  that  may 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     65 

be  used.     In  Fig.  47  a  profile  plane  is  used,  and  the  profile 
traces  yield  the  intersecting  point. 

In  Fig.  48  an  oblique  plane,  Q,  is  used,  which  intersects  T 
in  the  line  CD,  and  S  in  the  line  EF.  The  lines  CD  and  EF 
intersect  in  the  point  G,  through  which  the  line  of  intersection, 
AB,  may  be  drawn. 

Note. — The  foregoing  solutions  give  an  idea  of  the  multitude  of  resources 
open  to  the  student  in  the  solving  of  varieties  of  problems.  Sometimes  one 
method  or  auxiliary  plane  will  be  useful,  which  would  be  awkward  under 
other  conditions.  No  stereotyped  method  of  performing  these  solutions  can 
be  given,  for  each  problem  will  bring  up  conditions,  where  there  is  a  pref- 
erence of  methods.  The  analyses  are  exactly  alike,  and  they  must  be 
learned,  but  not  memorized. 


FIG.  48. 


EXERCISES 

66.  Draw  the  intersections  of  the  following  planes. 

282.  Q(l  +  3)  4(1  -  1)  and  R(5  +  3)  1(3  -  3). 

283.  8(1  +  3)  1(4  -  3)  and  T(l  +  2)  3(3  -  3). 

284.  U(5  +  3)  1(5  -  1)  and  W(3  +  3)  3(5  -  2). 

285.  Y(l  +  2)  4(1  -  1^)  and  Z(4  +  3)  1)4(1%  -  3). 

286.  X(  +  1)  oo  (  -  2)  and  Q(2  +  3)  5(1  -  2%). 

287.  R(  -  1)  «  (  +  3)  and  8(1  +  2)  3(1  -  1).      . 

288.  T(  +  1)  co  (  +  1)  and  U(l  +  1J£)  2(5  -  3). 

289.  W(  +  1)  oo  (  -  3)  and  Y(  +  2)  oo  (  -  2). 

290.  Z(  +  1)  » (  -  1M)  and  Q(  +  M)  «  (  ~  2)- 

291.  R(  -  2)  co  (  +  l)  and  T(  -  1)  co  (  -  2). 

292.  X(  -  3)  oo  (  +  1)  and  S(  +  2)  oo  (  -  3). 

293.  W(  +  2)  oo  (  -  1)  and  Z(  -  1)  » (  +  2). 

294.  X(l  +  3)  4(5  -  1)  and  Y(  +  2)  » (  -  %). 

295.  R(5  +  3)  3(1  -  3)  and  T(l  +  3)  4(5  -  1). 


66  PRACTICAL  DESCRIPTIVE  GEOMETRY 

296.  S(  +  IK,  parallel  to  tf)  and  X(5  +  2)  3(1  -  2). 

297.  U(  +  1,  parallel  to  7)  and  W(5  +  2)  3KU  -  3). 

298.  X(l  +  2)  3(1  -  1)  and  Y(5  +  IK)  3(5  -  3). 

299.  R(l  +  3)  3(1  -  2)  and  8(5  +  2)  3(5  +  1). 

300.  T(l  +  2)  3(5  -  2)  and  W(2  -  3)  3(4  +  3). 

301.  U(4  +  2)  3(1  -  1)  and  X(5  +  2)  4(5  -  2). 

302.  Y(l  +  3)  3(5  -  2)  and  Z(2  +  3)  4U  -  1). 

303.  R(l  +  3)  5(3  -  3)  and  Q(5  +  2)  3K(1K  -  3). 

304.  8(1  +  3)  2(1  -  3)  and  U(5  +  2)  3(5  -  3). 

305.  T(l  +  3)  2(5  -  1)  and  X(5  +  2)  4(1  -  2). 

306.  W(l  -  3)  2K(5  +  2)  and  Y(4K  -  3)  4(1  +  1). 

307.  Q(3  +  3)  1(2  -  3)  and  Z  (4  -  3)  5(3  -  3). 

308.  R(l  +  H)  (5  +  2K)  (1  -  H)  (5  -  IK)  and  8(1  +  3)  3(5  -  3). 

309.  T(2  +  3)  2(5  -  2)  and  U(5  +  2)  3(1  -  3). 

310.  W(5  -  3)  1(1  -  2)  and  X(l  -  2K)  3(5  +  1J4). 

311.  Y(3  +  3)  1K(4  -  2)  and  Z(5  +  2)  2(4K  -  2). 

312.  Q(l  +  2)  (5  +  3)  (1  -  1)  (5  -  IK)  and  R(l  +  1)  (5  +  IK)  (1  -  2) 
(5  -  3). 

313.  8(1  +  1)  (5  +  3)  (1  -  IK)  (3  -  2)  and  T(l  -  IK)  (3  -  2)  (1  -  1) 
(5  -  3). 

314.  U(l  +  2)  5(1  -  3)  and  a  30°  plane  which  contains  GL. 

314.  W(5  -  1)  1(4  +  3)  and  a  plane  containing  GL  and  A(3  +  1  -  2). 

315.  X(l  +  1)  (5  +  3)  (1  -  K)  (5  +  IK)  and  a  45°  plane  through  GL. 

316.  Draw  a  line  through  M(.4  +  1  -  1)  parallel  to  T(l  +  3)  4(1  -  1)  and 
8(5  +  3)  1(3  -  3). 

317.  Draw  a  line  2  in.  long  through  O,  (4  +  1  -  2)  parallel  to  W(l  +  3) 
2(5  -  1)  and  X(5  +  3)  3(1  -  IK)- 

318.  Draw  a  line  2K  in.  long  through  A (2  +3-3)  parallel  to  X(l  +  3) 
3(5  -  3)  and  Y(3  +  3)  2(1  -  3). 

TU9.  Find  the  point  of  intersection  of  the  three  planes,  R(l  +3)  5(1  —  2), 
N     S(  -  2)  co  (  -  K),  and  T(5  +  2)  1(3  -  3). 
320r  Draw  the  projections  of  the  tetrahedron  bounded  by  7£-and  the  three 

planes  in  Ex.  319. 

321.  Develop  the  surface  of  the  tetrahedron  in  Ex.  320. 
322\T)raw  the  projections  of  the  frustum  of  the  tetrahedron  in  Ex.  320 
\de  by  W(  +  K)  °°  (  -2). 

323.  Develop  the  surface  of  the  frustum  of  the  tetrahedron  in  Ex.  322. 

324.  Draw  the  projections  of  the  tetrahedron  formed  by  the  intersection 
of  the   planes   8(3  +  2)  K(5  +  IK),   T(3  -  3)  2(4  +  2),    R(3  -  3) 
4K(2  +  2%)  and  the  plane  Q,  2  in.  below  and  parallel  to  H. 

325.  Develop  the  surface  of  tetrahedron  in  Ex.  324. 

326.  Draw  the  projections  of  the  frustum  of  the  tetrahedron  in  Ex.  324, 
made  by  the  plane  Y(  -  1%)  « (  +  3). 

327.  Develop  the  surface  of  the  frustum  in  Ex.  326. 

THE  PIERCING  POINT  OF  A  LINE  AND  PLANE 

67.  Problem  18. — To  find  the  point  where  an  oblique  plane  is 
pierced  by  a  line. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     67 


^ 


V 


A 


<: 


7    / 

Or/ 


^ 


HXHtoV 


10 


VT 


\ 


VM 
HM 


14  V  / 

\t^/ 

-.*r^     l^!»«. 


15 


" 


\JS^ 

„**>  x, 


17  \   HU 


HUHtoV 


H5 


HT 
VT 


HX 


HY 


W  'Contains  GL  and  0 


Graphic  Layout  No.  23. — Draw  the  projections  of  the  line  of  intersection  of 
the  plaijes  given  in  Ex. . 


68 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


FIRST  METHOD 

Construction. — Let  T  (Fig.  49)  be  the  given  plane  and  AB  the 
given  line. 

1.  Obtain  QT. 

2.  Obtain  aqbQ. 

3.  Locate  OQ  (the  piercing  point)  where  aQbQ  intersects  QT, 
and  project  it  back  to  0HOV. 

Conclusion. — The 'point  0  is  the  required  piercing  point. 


FIG.  49. 


SECOND  METHOD 

Analysis. — 1.  Pass  a  plane  through  the  given  line. 

2.  Find  the  line  of  intersection  of  the  auxiliary  and  given 
planes. 

Conclusion. — Where  this  intersection  crosses  the  given  line  is 
the  piercing  point. 

Proof. — Since  the  point  thus  found  is  not  only  on  the  given 
line,  but  in  the  given  plane,  it  must  be  the  common  point  of  the 
two,  i.e.,  the  "piercing  point." 

Construction. — Let  AB  (Fig.  50)  be  the  given  line  and  T  the 
given  plane. 

1.  Pass  a  plane  (Q)  through  AB. 

Note. — While  any  of  the  infinite  number  of  planes  containing  AB  may 
be  used,  it  is  simpler  and  preferable  to  use  the  H-  or  F-projecting  plane.  In 
this  construction  the  //-projecting  plane  (Q)  is  used. 

2.  Obtain  MN,  the  intersection  of  Q  and  T. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     69 

3.  Locate  C,  where  MN  crosses  AB. 

Conclusion. — C  is  the  point  in  which  AB  pierces  T. 

Note. — This  problem  figures  in  most  of  the  succeeding  problems,  and 
should  be  thoroughly  mastered. 


Fio.  50. 


FIG.  51. 


70  PRACTICAL  DESCRIPTIVE  GEOMETRY 

68.  Problem  19. — To  solve  Problem  18  without  the  traces  of 
the  given  plane. 

Construction. — Let  the  two  intersecting  lines  MN  and  OP 
determine  the  given  plane,  and  AB  be  the  given  line  (Fig.  51). 

1.  Pass  the  projecting  plane  Q  through  AB. 
(Note. — HQ  is  sufficient.) 

2.  Project  the  Q-piercing  points  of  MN  and  OP,  CH  and  dH, 
to  their  F-projections,  cv  and  dv. 

3.  Draw  cvdv,  the  intersection  of  Q  and  the  plane  MNOP. 

4.  Where  cvdv  crosses  avbv,  at  xv,  the  piercing  point  is  located. 
Project  to  XH  on  aHbH. 

Note. — This  problem  is  of  great  value  in  surface  intersections. 

EXERCISES 

69.  Find  the  piercing  points  of  the  following  lines  and  planes.     Familiarize 
yourself  with  both  methods  by  doing  the  exercises  both  ways. 

328.  A(2  +  3  -  2)  B(4  +  1  -  1)  and  S(5  +  3)  1(5  -  iy2). 

329.  C(l  +  1M  -  M)  D(4  +  1M  -  K)  and  S  in  Ex.  328. 

330.  D(3  +  K  -  IK)  E(5  +  2  -  2)  and  S  in  Ex.  328. 

331.  F(3  +  2,  0)  G(3  +  2  -  3)  and  S  in  Ex.  328. 

332.  K(3  +  K  +  y^  L(4  +  3  +  3)  and  S  in  Ex.  328. 

333.  M(3  -  1  +  1)  N(4  -  2  +  3)  and  T(2  -  2)  5(3  +  3). 

334.  O(l  -  1  +  2)  P(5  -  1  +  2)  and  T  in  Ex.  333. 

335.  A(l  +  M  +  IK)  B(5  +  Yz  +  I'M)  and  T  in  Ex.  333. 

336.  C(3K  -2+3)  D(3K  -  %  +  K)  and  T  in  Ex.  333. 

337.  E(3>£  +  1  +  3M)  F(5  +  2  +  1)  and  T  in  Ex.  333. 

338.  G(2  +  2  -  1)  K(4  +  1  -  3)  and  X(5  +  3)  3(1  -  3). 

339.  M(l  +  2  -  1)  N(4  +2-1)  and  X,  Ex.  338. 

340.  O(4  +  3  -  1)  P(2  -  3  +  1)  and  X,  Ex.  338. 

.  341.  A(3  -  3  +  2)  B(l  -  1  +  1)  and  ¥(-!)«(+  2). 

342.  C(l  -  2  +  3)  D(3  -  K  +  K)  and  Y,  Ex.  341. 

343.  E(3  +  2  -  2)  F(3  +  %  -  1%)  and  U(  +  1%)  »(  -  1). 

344.  G(l  +  2  -  2)  K(3>£  +  1  -  %)  and  TJ(Ex.  343). 

345.  M(2  +  2  -  1)  N(3  +  1  -  IK)  and  R(5  +  2)  1(5  +  1). 

346.  O(3  +  3  -  K)  P(3  +  1  -  2)  and  R(5  +  2)  1(5  +  1). 

347.  A(l  +  1  -  1)  B(5  +  1  -  1)  and  R  (Ex.  346). 

348.  C(3  -  1  +  2)  D(5,  0  +  3)  and  R(Ex.  346). 

349.  E(2  +  3  -  3)  F(2  +  K  -  IK)  and  T(5  +2)  (1  +  1)  (5  -  3)  (1  -  IK). 

350.  G(2  +  2  -  Yri  K(3  +  K  -  IK)  and  T,  Ex.  349. 

351.  M(l  +  2  -  }4)  N(5  +  2  -  %)  and  T,  Ex.  349. 

352.  A(4  +  2  -  1)  B(3  +  1  -  3)  and  S(l  +  3)  2(5  -  2). 

353.  C(l  +  1  -  2)  D(4  +  1  -  2)  and  S,  Ex.  352. 
354    E(2  +  2  -  y2}  F(2  +  1  -  2)  and  S,  Ex.  352. 
355.  G(3  +  2  -  \Y2)  K(3%,  0-2)  and  S,  Ex.  352. 

Locate  the  piercing  points  of  the  following  lines  with  the  planes  of  the 
various  given  lines,  without  drawing  the  traces  of  the  planes. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     71 


356.  M(2  +2-1)    N(3M  +  1  ~  2^)    and   the   plane   of   O(5  +  3  -  1) 

y2  -  2). 


y^r 


7aa 

•I 
I 


10 


Given   Lines 


^ 


x 


VY 
HY 


z 


Tf 


10 


12 


PlaiN  Contains  Oand6L 


•i 


Given  Planes 

Graphic  Layout  No.  24. — Locate  the  point  in  which  the  line pierces 

the  plane .    Note. — Draw  the  line  so  that  it  is  likely  to  pierce  the  plane. 

The  H  and  V  projections  and  traces  may  be  transposed,  if  desired. 

357.  A(5  -  M  +  3)    B(2  -2^+1)    and    the    plane    of    C(5  -  2  +  2) 
D(l  -  1  +  1^)  E(5,  0  +  1). 


72 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


358.  P(3  +  2  —  3)    O(3  +  1  -  M)    and    the    plane    of    G(3  +2-2) 
K(l  +  1  -  iy2)  L(5  +  1,  0). 

359.  M(2  -3+2)    N(2  -  Y2  +  1)    and    the    plane    of    A(l  -  2  +  2K) 
B(4  -  1  +  1)  C(l  -  %  +  IK). 

360.  D(l  +  3  -  1)    E(3K  +  1-2)   and   the   plane   of   F(2  +  1^-3) 
G(4  +  2M  -  1)  and  K(2^  +  K  -  2)  L(4^  +  1%,  0).    ' 

361.  M(3  +  2  -  K)  N(3  +  K  -IK)  and  the  plane  of  FG  -  KL,  Ex. 
360. 


H 


Graphic  Layout  No.  26.  —  Find  where  the  given  line  pierces  the  given  plane 
in  Ex.  -  . 


362.  O(4  -  2  +  Y2)  P(2^  -  H  +  2K)  and  the  plane  of  A(2  -  1^,  0) 
B(4  -  YZ  +  2)  and  C(2  -2^+1)  D(4  -  IK  +  3). 

363.  From   the  middle  point  A  of  the  line  B(l  +  2  -  1)  C(l  +  1  -  3) 
draw  a  line  which  shall  intersect  M(2+H  —  1)  N(4  +  3  —  3)  and 
0(3M  +  IK  -  1)  P(4  +  M  - 


Graphic  Layout  No.  26.  —  Find  where  the  given  line  AB  pierces  the  plane 
of  the  intersecting  or  parallel  lines  in  Ex.  -  ,  without  drawing  the  plane- 
traces. 


364.  Draw  a  right  hexagonal  prism,  its  center  line  E(l  +  2£  —  1) 

%  —  2),  with  one  of  its  faces  in  H.     Find  the  points  in  which  it  is 
pierced  by  G(1H  +  2  -  H)  K(3,  0  -  2%). 

365.  Draw  the  projections  of  the  shortest  line  on  the  surface  of  the  hexagonal 
prism  (Ex.  364)  between  the  piercing  points  of  GK. 

366.  Hexagonal  prism  in  Ex.  364.     Find  the  piercing  points  of  the  three 
parallel  lines  A(l,  0  -  2)  B(3  +  2  -  2),  C(1M,  0-1)  D(3K  +  2  - 
1),  and  E(2,  0  -  IK)  F(4  +  2  - 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     73 


367.  Hexagonal  prism  in  Ex.  364.  Find  the  piercing  points  of  the  three 
parallel  lines  G(l^,  0-1)  K(l^  +2-1),  L(2,  0-2)  M(2  +  2  - 
2),  and  N(2M,  0-1)  P(2^  +2-1). 

368  Draw  a  right  square  pyramid  whose  base  diagonal  is  A  (2,  0  —  1) 
B(4,  0-2)  and  apex  C(3  +  2%  -  !>£)•  Find  the  points  in  which 
it  is  pierced  by  the  line  E(1J£,  0  -  2)  F(4  +  2  -  ^). 

369.  Pyramid  in  Ex.  368.     Find  the  points  in  which  it  is  pierced  by  four 
lines  parallel  to  GL,  through  the  points  M(4  +  1  -  1%),  N(4  +  %  - 
IK),  0(4  +  1_-  %),  and  P(4  +  \Y2  -  1^)- 

370.  Pyramid  in  Ex.  368.     Find  the  points  in  which  it  is  pierced  by  four 
lines  ||  to  V  and  45°  to  H,  through  the  points  M,  N,  O  and  P  in  Ex. 
369. 

371.  Pyramid  in  Ex.  368.     Find  its  intersection  with  the  solid  whose  four 
edges  are  the  lines  in  Ex.  370. 

372.  Pyramid  in  Ex.  368.     Find  its  intersection  with  the  prism  whose  four 
edges  are  the  lines  in  Ex.  369. 


Graphic  Layout  No.  27.- 
solid  in  Ex.  . 

(6)  Find  the  intersection  of  solid  in  Ex. 
ing  plane  of  AB.     Find  its  true  size. 

(c)  Pass  a  plane  through  solid  in  Ex.  - 
projections  of  AB,  Ex. 


-(a)  Find  where  the  given  line  pierces  the  given 


by  the 


(H,  V)  project- 


-,  its  traces  |j  to  the  respective 
Find  the  intersection  and  true  size, 
in.  to  the  right  of  the  center  of  solid .    Find 


(d)  Pass  a  profile  plane 

the  intersection  and  true  size. 

(e)  Develop  the  truncated  portion  of  the  solid  made  by  the  plane  in  one 
of  the  foregoing  exercises. 

373.  Pyramid  in  Ex.  368.     Find  its  intersection  with  the  plane  T(5  +2) 
K(3  —  3).     Crosshatch  the  section  and  show  it  in  its  true  size. 

374.  Develop  the  surface  of  the  pyramid  in  Ex.  373,  between  H  and  T. 
Make  a  paper  model  of  it. 

375.  Pyramid  in  Ex.  368.     Find  its  intersection  with  the  plane  S(  + 
°°  (  —  3).     Crosshatch  the  section  and  show  it  in  its  true  size. 


74 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


376.  Develop  the  surface  of  the  pyramid  in  Ex.  375  between  H  and  S. 
Make  a  paper  model  of  it. 

377.  Prism  in  Ex.  364.     Find  its  intersection  with  the  plane  U(2  +  3)  2- 
(3  —  3).     Crosshatch  the  section  and  show  it  in  its  true  size. 

378.  Develop  the  surface  of  the  prism  in  Ex.  377  between  one  of  its  bases 
and  U.     Make  a  paper  model  of  it. 

379.  Prism  in  Ex.  364.     Find  its  intersection,  and  the  true  size  of  the  inter- 
section, with  a  profile  plane  through  A(2^,  0,  0). 

380.  Develop  the  surface  of  the  prism  in  Ex.  379  between  one  base  and  the 
profile  plane.     Make  a  paper  model. 

381.  Prism  in  Ex.  364.     Draw  its  intersection  by  the  plane  S(2  +  3)  4^- 
(1  —  3).     Crosshatch  the  intersection  and  show  its  true  size. 

382.  Develop  the  prism  in  Ex.  381  between  one  of  its  bases  and  S.     Make 
a  paper  model. 

PERPENDICULARS  TO  PLANES 

70.  Problem  20. — To  draw  a  line  perpendicular  to  a  given 
plane,  through  a  given  point. 


FIG.  52. 

Analysis. — 1.  Through  the  ^-projection  of  the  given  point 
draw  the  F-projection  of  the  required  line  perpendicular  to  the 
F-trace  of  the  given  plane. 

2.  Through  the  //-projection  of  the  given  point  draw  the  H- 
projection  of  the  required  line  perpendicular  to  the  //-trace  of 
the  given  plane. 

Construction. — Let  the  student  make  the  construction  for  the 
ordinary  case. 

Special  Case. — When  the  given  plane  is  parallel  to  GL. 

Construction. — Let  M  (Fig.  52)  and  T  be  the  given  point  and 
plane. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     75 

1.  Obtain  the  P-projection  of  M. 

2.  Obtain  the  P-trace  of  T. 

3.  Draw  the  perpendicular  mpnp  to  PT. 

4.  Project  np  to  nv  and  nH  on  a  line  through  M  perpendicular 
to  the  traces  of  T. 

Conclusion. — MN  is  perpendicular  to  T. 

Note. — N  need  not  be  a  point  in  the  plane,  but  any  point  in  the  line. 


71. 


EXERCISES 


HU 


vu 


\ 


VK 
HK 


VM 
HM 


10 


VN 


UN 


Graphic  Layout  No.  28.— From  A (3  +  2  -  2)  (or  assume  the  point,  if 
preferred)  draw  a  perpendicular  to  plane  .  Find  where  the  perpen- 
dicular pierces  the  given  plane.  Find  the  true  length  of  the  perpendicular 
from  the  point  to  the  plane. 

383.  From  A (2  +  2  -  2^)  draw  a  perpendicular  to  T(5  +  2)  1(4  -  3), 
and  find  where  it  pierces  it. 

384.  From  B(l  -  2  +  1)  draw  a  perpendicular  to  S(5  -  3)  5(2  +  3),  and 
find  where  it  pierces  it. 

385.  From  C(4  -  1  -  2)  draw  a  perpendicular  to  R(3  -  3)  1(4  -  3),  and 
find  where  it  pierces  it. 

386.  From   D(2  +  1   -  H)  draw   a   perpendicular  to  X(  +2)o>(  -  2), 
and  find  where  it  pierces  it. 


76 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


387.  From   E(2  -  1  +  K)   draw   a   perpendicular  to   U(5  —  2)   (1  —  1) 
(5+3)  (1  —  1H)>  and  find  where  it  pierces  it. 

388.  From  F(3  +  2  -  2)  draw  a  perpendicular  to  W(l  +  2)  3(5  -  2),  and 
find  its  W-piercing  point. 

389.  From  G(3,  0,  0)  draw  a  perpendicular  to  Y(l  +  2)  4(1  —  3),  and  find 
where  it  pierces  it. 

390.  From  K(3,  0,  0)  draw  a  perpendicular  to  T(  —  2)  <=o  (  -  1). 

391.  Scale  ^  in.  =  1  ft.     The  roof  of  a  building  contains  two  points:  A 
(16  ft.,  0,  -  8  ft.),  and  B(22  ft.,  +  14  ft.  6  in.,  -  2  ft.).     The  ridge 
and  lower  edge  run  24  ft.  horizontally  to  the  southeast.     At  C(28  ft., 
0  —  6  ft.)  it  is  desired  to  erect  a  rafter  brace  perpendicular  to  the  roof. 
At  what  point  will  it  support  the  roof,  and  how  long  will  it  be? 

392.  Scale  J^  in.  =  1  ft.     A  tower,  a  hexagonal  pyramid,  18-ft.  base,  16-ft. 
altitude,  base  center  at  0(24  ft.,  0-8  ft.).     From  B(42  ft.  +  20  ft.  - 
18  ft.)  it  is  desired  to  run  a  telephone  wire  perpendicularly  to  the 
nearest  face.     Where  is  the  hole  drilled,  and  what  is  the  length  of  the 
wire? 

72.  Problem  21. — To  erect  a  perpendicular  of  any  given  length 
from  a  point  in  a  plane. 


JL 


FIG.  53. 


Construction. — Let  A  (Fig.  53)  be  the  given  point,  and  T  the 
given  plane.     Required  to  erect  a  perpendicular  1  in.  long. 

1.  Draw  a  perpendicular  to  T  of  indefinite  length  through  A. 

2.  Obtain  QT. 

3.  Obtain  aQ- 

4.  Erect  the  perpendicular  aQcQ,  1  in.  long. 

5.  Project  CQ  back  to  CH,  thence  to  cv. 

Conclusion, — AC  is  perpendicular  to  T,  and  1  in.  long. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES    77 

SECOND  METHOD 

Analysis. — 1.  Erect  a  perpendicular  of  indefinite  length  from 
the  given  point,  according  to  Problem  20. 

2.  Assume  a  point  on  the  perpendicular. 

3.  Revolve  the  definite  line,  thus  drawn,  so  as  to  show  its  true 
length. 

4.  Lay  off  the  required  distance  on  the  true  length  of  the  line 
from  the  given  point. 

5.  Return  the  limiting  point  to  its  proper  position  in  space. 


Construction. — To  erect  a  perpendicular  1  in.  long  at  A  (Fig. 
54)  in  the  plane  T. 

1.  Draw  aHbH  and  avbv,  respectively  perpendicular  to  HT  and 
VT. 

2.  Having  assumed  B  on  the  line,  revolve  it  parallel  to  V,  to 
show  its  true  length. 

3.  Lay  off  1  in.  on  this  revolved  line  avciv. 

4.  Return  dv  to  CVCH  on  AB. 

Conclusion. — AC  is  perpendicular  to  T,  and  1  in.  long. 


78 


74. 


PRACTICAL  DESCRIPTIVE  GEOMETRY 
EXERCISES 


393.  From  A (2  +  1,  x)  in  T(l  +  3)  4(1  -  2)  erect  a  perpendicular  2  in. 
long. 

394.  From  B(l,  y  —  1)  in  S(  +  2)  <»(  —  1)  erect  a  perpendicular  \y%  in. 
long. 

395.  From  C(2,  z  -  1)  in  R(l  +  2^£)  5(1  +  1)  erect  a  perpendicular  2  in. 
long. 

396.  From  D(3  -  1,  y)  in  Q(l  -  3)   (5  -  1)  (1  +  \Y^>  (5  +  K)  erect  a 
perpendicular  \%  in.  long. 


1  */ 

2     V 

3  HU  

'    #x 

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y 

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/ 

s--ja^ 

6v 

7 

8 

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VK 
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9        VM 
HM 

10 
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in.  perpendicular  to  the  plane  from  the  assumed  point. 

397.  From  E(2  -  1%,  x)  in  U(2  -  3)  3^(5  +  3)  erect  a  perpendicular  1  in. 
long. 

398.  Draw  the  projections  of  a  right  prism,  1%-in.  altitude,  having  M(l  +  2 

-  11A)  O(l%  +  1-2)  P(2^  +  \Y2  -  1)  as  its  base. 

399.  Assume  the  projections  of  any  plane  figure  in  space  and  erect  a  right 
prism  of  2-in.  altitude  on  it  as  a  base. 


THE  PROJECTIONS  OF  OBLIQUE  SOLIDS 

76.  Problem  22. — To  draw  the  projections  of  a  solid  having 
its  base  in  an  oblique  plane. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     79 


FIG.  55. 


<J, 


FIG.  56. 


80 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


FIRST  METHOD 

Construction. — Let  T  (Fig.  55)  be  the  given  plane,  arid  let  it 
be  required  to  draw  a  1-in.  X  1^-in.  hexagonal  prism. 

1.  Obtain  QT.' 

2.  With  QT  as  a  ground  line  draw  the  solid  in  its  natural 
position. 

3.  Project  it  back  on  H  and  V,  following  the  dotted  projectors, 
as  shown  in  the  figure. 


ICt 


FIG.  57. 


76. 


SECOND  METHOD 


Construction. — Let  0  (Fig.  56)  be  the  center  of  a  1-in.  square 
in  the  plane  T.  Let  it  be  required  to  draw  a  pyramid  of  2-in. 
altitude  on  this  square  as  a  base. 

1.  Revolve  0  about  HT  into  H. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     81 

2.  Construct  a  1-in.  square  about  O,  as  a  center. 

3.  Counter-revolve  the  square  into  T,  as  in  Problem  16. 

4.  At  0  erect  OM,  a  2-in.  perpendicular  to  T. 

5.  Connect  M  with  the  four  corners  of  the  square. 

77.  Supplementary  Problem. — To  draw  the  projections  of  a 
solid,  having  its  center  line  given. 

Construction. — Let  AB  (Fig.  57)  be  the  center  line  of  an  equi- 
lateral triangular  prism,  faces  1  in.  X  3  in.  altitude. 

1.  Locate  the  Q-projection  of  AB. 

2.  Draw  QX  perpendicular  to  AB,  locating  axbx,  a  point. 

3.  Draw  the  Q-  and  X-projections  of  the  prism. 

4.  Project  these  views  on  H,  and  thence  on  V. 


78. 


EXERCISES 


In  the  following  exercises,  twelve  typical  solids  are  given,  to  be  erected  with 
their  bases  in  the  given  planes  with  their  base  centers  at  given  points  in  the 
planes.  Where  the  base  diameter  is  given  it  signifies  the  diameter  of  the 
circumscribing  circle.  Consider  the  solids  opaque,  and  dot  the  invisible 
lines.  (See  Figs.  55,  56  and  57.) 


(I) 


OQO 


(4)  (5) 


(6) 


The  solids  to  be  erected : 

(a)  Equilateral  triangular  prism,  2-in.  altitude. 

(6)  Equilateral  triangular  pyramid,  2-in.  altitude. 

(c)  Square  prism,  2-in.  altitude. 

(d)  Square  pyramid,  2-in.  altitude. 

(e)  Rectangular  prism,  2-in.  altitude. 

(/)  Rectangular  pyramid,  1%-in.  altitude. 

(g)  Pentagonal  prism,  2-in.  altitude. 

(k)  Pentagonal  pyramid,  2-in.  altitude. 

(Z)  Hexagonal  prism,  2-in.  altitude.    • 

(m)  Hexagonal  pyramid,  2-in.  altitude. 

(n)  Circular  cylinder,  2-in.  altitude, 

(o)  Circular  cone,  2-in.  altitude. 

Place  one  of  the  foregoing  solids  ("the  instructor's  assignment)  on  the  fol- 
lowing planes,  the  center  at  the  points  given,  and  the  base  diagonal  parallel 
to  H  or  V,  according  to  assignment. 


400.  Solid 
6 


-,  Q(l  +  3)  1(4  -  3),  Center  O(3  +  1,  x). 


82 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


401.  Solid 

402.  Solid 

403.  Solid 

404.  Solid 

405.  Solid 

406.  Solid 

407.  Solid 

408.  Solid 

409.  Solid 

410.  Solid 

411.  Solid 

412.  Solid 

413.  Solid 


R(l  -3)  1(5+2),  A(4  -1,  y). 

S(4  +  2)  2(4  -  1),  B(4,  x  -  H). 

T(l  -  3)  3(1  +  2),  CUM  -  1,  x). 

U(l  +  2)  2^(4  -  2),  D(2M,  y  -  1). 

W(5K  -  2)  4(2K  +  2),  E(3K  -  %,  x). 

X(4  +  3)  1(5  +  1),  F(3+y  -  M). 

Y(l  +  IK)  (5  +  %)(!-  1)  (5  -  M),  G(2  +  1,  x). 

Z(5  -  3)  (1  -  2)  (5  +  2M)  (1  +  IK),  K(3  -  1%, 

8(1  +  3)  2(5  -  IK),  L(4  +  2,  x). 

T(5  -  2K)  4(1  +  1),  M(2,  y  +  IK). 

U(+2)«(  -  1),  N(3  +  1,  x). 

W(  -3)  »(  +  i),  0(3  -1,  y). 

X(  +  l)-(  -M),  P(2,x  -2). 


Erect  one  of  the  foregoing  solids  on  one  of  the  following  center  lines. 
Draw  one  side  of  the  base  ||  to  H  or  V,  according  to  assignment. 

1)  B(3  +2  -  1). 

+  1M)  D(l  -  1  +  IK). 
F(3  -  2  +  IK)- 

2)  K(3  -  1  +  I). 

N(2  +  K  -  2M). 
1)  P(5  -  1  +  2K)- 

3)  B(2  +  1  +  M). 
1)  D(5  -  2K  +  3). 
K)  F(2  +  1  -  2). 

-  2K)  K(2  +  1  -  1). 
N(2  -  %  +  2). 
(2  -H+  2K). 

Erect  solid  b,  d,  f,  k,  m,  or  o,  according  to  assignment,  with  its  apex  in  the 
point  given,  and  its  base  in  the  plane  given  in  the  following  exercises.  Disre- 
gard the  altitudes  given,  as  they  may  not  fit. 


• 

r^or>foT-   lino    P  fO  I/              1 

416.  Solid  - 

—  ,  Center  line  E(l  -  1  + 

.    rVmtnr  linn  P.  (*\           1.1 

fVn+pr  lino  TVf  H    4-9 

PnntnT-   linn    C\(  A              1       1 

4.9O      ^nlirl 

Pon  +  oi-   lino     A  f  1        I      O 

491        Qnlirl 

frmtnr  linn   ("'('A           1     .1 

4.99      Snlirl 

Pnnfnr  linA  TT/O      1     9 

4.93      Sr\lir1 

PpntAr  linf>   <~i^9    4-  91,^ 

4.94.      Qrtlirl 

4.9  *      SrOirl 

Pan+OT-  lino    ^("9              1     -U 

426.  Solid 

427.  Solid 

428.  Solid 

429.  Solid 

430.  Solid 

431.  Solid 

432.  Solid 

433.  Solid 
(5  - 

434.  Solid 
(5  - 

435.  Solid 

436.  Solid 

437.  Solid 

438.  Solid 

439.  Solid 

440.  Erect 
O(2  - 


Apex  A(5  +  IK  -  M),  Base  T(l  +  3)  1(4  -  2). 
Apex  B(3  +  2  -  2),  Base  S(l  +  2)  2K(2K  -  3). 
Apex  C(l  +  3  -  2),  Base  TJ(2+  2)  5(2-  1). 
Apex  D(3  -  2K  +  3),  Base  W(5  -  2)  3(5  +  3). 
Apex  E(l  +  2  -  K),  Base  X  (5  +  2)  4(1  -  1). 
Apex  F(3  -  1  +  2),  Base  Y(l  -  3)  2(5  +  1). 
Apex  G(3  +  1  -  2H),  Base  Z(5  +  3)  1(5  +  1). 
Apex  K(3,    0,    0),     Base    Q(l  +3)     (5  +  2)     (1  - 


-,  Apex  M(4  -  3  +  2M),  Base  R(l  -  IK) (5  -  H)  (1  -  2) 


1). 

,  Apex  N(5  +  2  -  K),  Base  S(l  +  3)  3(5  -  3). 

— ,  Apex  O(l  -  3  +  1),  Base  T(5  -  2)  3K(2  +  2). 

— ,  Apex  P(3  +  3  -  3),  Base  U(  +  2)  « (  -  1). 

— ,  Apex  A (2  -  2  +  2),  Base  W(  -  2M)  «(  +  !)• 

— ,  Apex  B(2  -  2  +  1),  Base  X(  +  K)  °°(  +  1M)- 
a   1-in.   cube  on  T(l  +  2)   4(1  -  1),  with  its  base  center  at 
-  1,  x),  diagonal  ||  to  H. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     83 


441.  Draw  a  1-in.  cube  on  S(  +  2)  °°  (  —  1)  with  one  corner  at  A(3  +  Y±,  x) 
and  another  touching  H. 

442.  Draw  a  cube,  one  of  whose  edges  is  the  line  A  (4  +  1  —  2)  B(3  +  1—1) 
and  another  touches  H. 

443.  The  center  line  of  a  cube  is  C(2  +  2  -  2)  D(3  +  1  -!}$)•     Draw 
the  cube,  (a)  with  an  edge  ||  to  H;  (6)  with  its  base  diagonal  ||  to  H. 

444.  The  line  E(2  +  11A,  *)  F(3  +  %,  y)  in  S(  +  2)  °°  (  -  i^)  is  one  edge 
of  a  cube,  base  in  S.     Draw  the  projections  of  the  cube. 


/''  ' 

(I) 


ooo 

l,.../^--.)    U-,/j''-.>l    U-//-f-4 
(4)  (5)  (6) 


HX 


HW 


8 


10 


12      . 


° 


Graphic  Layout  No.  30. — Let  the  polygons  be  bases  of  2-in.  prisms  or 
pyramids  (or  cones  or  cylinders  for  No.  6). 

(a)  Erect  prism  (or  cylinder),  base  on  plane ,  assuming  a 

center  in  the  plane. 

(6)  Erect  pyramid  (or  cone),  base  on  plane ,  assuming  a 

center  in  the  plane. 

(c)  With  the  base on  plane under  any  of  the  conditions  given 

in  Graphic  Layout  No.  22,  erect  a  2-in.  pyramid  or  prism. 

(d)  Assume  a  convenient  point  in  space  (e.g.,  A(3  +  2  —  2)  or  B(3  — 

2  +  2))  as  the  apex  of  pyramid ,  or  cone ,  to  be  drawn  with  its 

base  in  plane . 

(e)  Draw  prism,  base  ,  with  its  center  line  parallel  to (H,  V, 

or  P),  one  of  its  faces  touching  plane . 

(/)  Draw  hollow  prism  or  cylinder,  with  Y^-va..  walls,  under  the  conditions 
(a)  or  (e). 


84 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


445.  The  lines  G(2  +  1  -  1)  K(l  +  K  -  1M)  and  KL(2^  +  1^ 

form  two  sides  of  a  parallelogram,  which  is  the  base  of  a  prism  of 
1^-in  altitude.     Draw  the  prism. 

446.  The  line  M(l  +  1  -  1)  N(2  +  K  -  %)  is  one  side  of  a  rectangle 
which  forms  the  base  of  a  prism  1^-in.  altitude.     The  opposite  side 
passes  through  D(2  +  1  —  1).     Draw  the  prism. 

447.  Draw  a  line  from  P(2>£  +  2  -  K)  that  shall  intersect  M(l  +  2  - 

N(3  +  H  -  1)  and  be  at  a  distance  of  2  in.  from  P(3^  +  1  - 


9H 


loM 


•  H 


(1) 


(5) 


(6) 


(3)  (4) 

Center  lines  for  solids. 

Graphic  Layout  No.  31. — (a)  With  line as  center  line,  draw 

prism  (or  cylinder)  with  base .     (6)  With  line as  center  line, 

a  2-in.  pyramid  (or  cone)  with  base . 

Note. — The  signs  on  these  lines  may  be  transposed. 


2-in. 
draw 


79.  Problem  23. — To  pass  a  plane  through  a  given  point  per- 
pendicular to  a  given  line. 

Principle. — The  traces  of  a  plane  perpendicular  to  any  line  are 
respectively  perpendicular  to  the  projections  of  the  line;  hence 
we  know  the  direction  of  the  traces. 

Analysis. — Apply  the  principle  given. 

Construction. — Let  0  (Fig.  53)  be  the  given  point,  and  AB  the 
given  line. 

1.  Draw  oHpH  perpendicular  to  aHbH. 

2.  Draw  ovpv  parallel  to  GL. 

3.  Obtain  Q  (P)  of  OP. 

4.  Through  0  draw  VT  perpendicular  to  avbv. 

5.  Where    VT    intersects    GL    draw    HT,    perpendicular    to 
aHbH. 

Conclusion. — T  contains  O,  and  is  perpendicular  to  AB. 

Notes. — 1.  If  the  first  trace  does  not  intersect  GL  within  the  problem 
limits,  draw  a  second  auxiliary  line  parallel  to  the  second  required  trace. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     85 

2.  If  the  given  line  lies  in  a  profile  plane,  the  simplest  solution  will  be 
accomplished  through  the  use  of  the  P-projection. 


80. 


EXERCISES 


—  1)  pass  a  plane  perpendicular  to  B(l  +  1  —  * 


448.  Through  A(3  + 
C(2  +  3  -  2). 

449.  Through  D(4  -1+1)  pass  a  plane  J.  to  E(2  -  3  +  1)  F(5  -  1  +  2). 

450.  Through  G(2  -  1  +  1)  pass  a  plane  _L  to  K(4  +  1  -  1)  L(5  +  3  -  2). 

451.  Through  M(3  +  1-1)  pass  a  plane  J_  to  N(4>i  +  %  -  1)  P(5  + 
3  -  2M). 

452.  Through    O(2  -  1  +  2)  pass   a    plane    _L  to  A  (3  +  2  -  M)   B(3  + 
1  -3). 


FIG.  58. 


453.  Through  P(2  +  \Y2  -  2)  pass  a  plane   _|_  to  C(3  +  2  +  1)  D(3  - 

1+M). 

454.  Through  K(5  +  1-1)  pass  a  plane  J_  to  E(3  +  3  -  1)  F(2  +  1  -  3). 

455.  Pass  a  plane  _L  to  G(2  +  1  -  1)  K(3  +2-3)  through  the  middle 
point  of  the  line. 

456.  Pass    a   plane  _L  to    M(l  +  1  -  ^)    N(4  +  3  -  1^)    through    the 
point  M. 

457.  One  edge  of  a  cube  is  the  line  A(3  +  11A  -  1%)  B(4  +  %  -  1)  and 
one  of  its  corners  touches  H.     Draw  its  projections. 

458.  M(3H  +  2  -  2%)  N(l  +  1^  -  1^)  is  the  base  of  an  isosceles  tri- 
angle whose  vertex  is  in  GL.     Draw  its  projections  and  revolve  to 
show  its  true  size. 


86 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


459.  The  center  line  of  a  stick  1%  in.  square  is  the  line  O(2  +'3  -  2)  P(3  + 
1  —  1).     Two  3^-in.  holes  are  drilled  straight  through  the  stick,  %  in. 
from  the  ends.     Draw  the  projections  of  the  stick  and  holes,  dotting 
invisible  lines. 

Note. — This  exercise  can  be  performed  by  the  method  of  the  Supple- 
mentary Problem,  Problem  22.  Conversely,  Ex.  414-425  can  also  be 
constructed  by  the  use  of  Problem  23. 

460.  The  upper  surface  of  a  vein  of  coal  is  found  to  be  perpendicular  to  a 
shaft  bored  in  the  direction  of  C(3  -  1  +  y^)  D(l  -3+2).     The  ore 
is  located  at  a  point  O(2,  x,  y)  on  CD.     Draw  the  traces  of  the  plane  of 
the  ore. 

461.  Scale  1  in.  =  10  ft.     A  4-ft.  stack  on  a  building  has  its  center  line  A(15  ft. 

—  30  ft.  +  15  ft.)  B(15  ft.,  O  +  15  ft.),  a  guy  wire  runs  from  a  point 
15  ft.  up  the  stack  to  a  point  C(23  ft.  -  22  ft.  +  7  ft.  6  in.)  in  a  roof 
perpendicular  to  the  wire.  Find  the  traces  of  the  roof  plane,  and  draw 
the  projections  of  the  roof,  a  rectangle,  40  ft.  long  by  20  ft.  slant  height, 
with  C  as  its  center. 

462.  Scale  1  in.  =6  ft.     A  3-in.  shaft  carrying  a  54-in.  pulley  (8-in.  face) 
mounted  7  ft.  from  the  upper  end,  runs  from  D(18  ft.  O  +  12  ft.) 
in  the  ceiling  of  a  factory  building  to  E(18  ft.,  18  ft.,  O)  in  a  wall. 
Draw  the  projections  of  the  shaft  and  pulley  (without  spokes). 

463.  Scale  1  in.  =  6  ft.     A  guide  pulley,  54-in.  X  12-in.  face,  is  running 
on  an  oblique  2-in.  shaft,  whose    center   line    is    F(12   ft.   —  6  ft. 
+  8  ft.  6  in.)  G(24  ft.  -  12  ft.  +  19  ft.  6  in.)     Draw  the  projections 
of  the  pulley  at  the  center  of  FG. 


Graphic  Layout  No.  32.  —  (a)  Assume  a  point  in  space,  and  pass  a  plane 
through  it  perpendicular  to  line  --  . 

(6)  Pass  a  plane  _l_  to  line  -  at  its  middle  point. 

(c)  Let  line  --  be  one  edge  of  a  cube,  with  one  corner  (or  edge)  touching 


—  (H  or  V).     Draw  the  projections  of  the  cube. 

(d)  Let  line  --  be  the  center  line  of  a  1^-in.  square  stick,  with  a  ^-in. 
hole  drilled  lengthwise  through  the  center.     Draw  its  projections. 

(e)  Let  line  --  be  the  center  line  of  a  1%-in.  square  stick  with  a  M-in. 
hole  drilled  crosswise  through  the  center.     Draw  its  projections. 

(/)  Let  line  --  be  the  center  line  of  a  2-in.  shaft,  carrying  a  24-in.  X  8-in. 
pulley.     Draw  the  projections  of  the  pulley  to  any  convenient  scale. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     87 

DISTANCE  OF  A  POINT  TO  A  PLANE 

81.  Problem  24. — To  find  the  distance  from  a  point  to  a  plane. 
Construction. — Let  0  (Fig.  59)  and  T  be  the  given  point  and 
plane. 

1.  Obtain  QT  and  OQ. 

2.  Draw  a  perpendicular  from  OQ  to  QT. 

Conclusion. — This  perpendicular  will  be  the  distance  required. 


FIG.  59. 


82. 


EXERCISES 


465. 
466. 
467. 

468. 
469. 
470. 
471. 
472. 
473. 

474. 


Shortest  distance  A(l  +  ^  -  2)  to  T(l  +  2)  4(4  -  3). 

Shortest  distance  B(l  -  3  +  2)  to  8(4  -  2)  1(1  +  3). 

Shortest  distance  C(2  +  1  -  3)  to  R(5  +  2)  2(5  -  2). 

Shortest  distance   D(2  +  %  -  K)   to   Q(l  +3)    (5  +  2)    (1  -  2H) 

(5  -  1H). 

Shortest  distance  E(l  +  H  -  3)  to  U(l  +  2)  3(5  -  2). 

Shortest  distance  F(4  -  1  +  1)  to  W(5  +  3)  2(5  -  1). 

Shortest  distance  M (4  -  2  +  2)  to  X(l  +  3)  2(5  -  1). 

Shortest  distance  N(2  +3  -  2)  to  Y(  +  1%)  «(  +  !). 

Shortest  distance  O(3  -  1  +  1)  to  Z  (+  1  »(  -  2). 

Shortest  distance  upper  right-hand  rear  corner  of  a  3-in.  cube  with 

faces  in  H  and  V,  to  the  plane  of  the  three  adjacent  corners. 

Scale  IJ^-in.  =  1  ft.     An  8-in.  circular  flue  runs  from  a  point  A(l  ft., 

0—8  in.)  in  a  floor  perpendicular  to  a  plane  whose  traces  intersect 


88  PRACTICAL  DESCRIPTIVE  GEOMETRY 

at  a  point  4  ft.  to  the  right,  the  F-trace  being  inclined  67^  to  GL,  and 
the  fl-trace  45°.  Draw  the  projections  of  the  roof  and  floor  openings 
for  the  flue. 


Graphic  Layout  No.  33. — Assume  a  point  in  space,  and  measure  its  dis- 
tance from  plane . 

PROJECTION  OF  LINES  ON  OBLIQUE  PLANES 

83.  Problem  25. — To  project  a  line  on  an  oblique  plane. 
Construction. — Let  AB  (Fig.  60)  be  projected  on  T. 

1.  Let  fall  perpendiculars  from  A  and  B  to  T. 

2.  Obtain  QT  and  aQbq. 

3.  Draw  the  projectors  aQmQ  and  bqnQ,  perpendicular  to  QT. 

4.  Project  the  points  mQ  and  nQ  back  to  the  H-  and  V-projec- 
tions  of  the  perpendiculars  drawn  in  operation  (1). 

Conclusion. — MN  is  the  projection  of  AB  on  T. 
Check. — If  AB  intersects  MN  at  the  point  in  which  AB  pierces 
T,  the  operation  is  probably  correct.     Why? 


84. 


EXERCISES 


475;  Project  A(1M  +  2  -  1)  B(3  +  1  -  2)  on  Q(5  +  3)  5(3  -  2). 

476.  Project  C(l  -  2  +  1)  D(2  -  %  +  2)  on  R(3  -  3)  4(4  +  3) 

477.  Project  E(2  +  1  -  1)  F(4  +  1  -  1)  on  8(1  +  3)  4(1  -  2). 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     89 


478.  Project  G(2  +  2  -  K)  K(3  +  1  -  3)  on  T(  +  2)  « (  -  1). 

479.  Project  L(3  -  1  +  1)  M(3  -  2  +  3)  on  U(  +  1)  « (  -  2>i). 

480.  Project  N(2  -  %  +  1)  P(4  -  1  +  2)  or  W(  -  1)  °°  (  +  2). 

481.  Project  G(3  +  2  -  1)  K(4^  +  1  -  3)  on  X(l  +  2)  3(5  -  2). 

482.  Project  A(l  +  1  -  1)  B(3  +  3  -  2)  on  Y(l  +  3)  (5  +  IK)  (1  -  2) 
(5  -  1). 

483.  Project  C(3  -  1  +  2)  D(4  -  2  +  Y2)  on  Z(l  +  3)  5(1  +  1). 

484.  Project  E(2  +  2  -  3)  F(4  +  1  -  1)  on  R(l  +  3)  1(5  -  1). 

Note. — Any  of  the  above  lines  may  be  projected  on  any  of  the  given 
planes. 


FIG.  60. 


17 


Graphic  Layout  No.  34. — Project  given  line  on  given  plane  in  Ex. . 


90 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


m  n 


i" 


lmH 


10 


V.  V 

g  k 


Given   Lines 


'^> 

y 

3 
^y^£, 

4 

A 

*\ 

\ 

HX 

5  
VY 

HY 

6 

7 

a 

\l7 

^ 

'-^sr 

"TTf  —  - 

Q 

10 

n 

fov 

1 

12 

c^- 

loM 
Plan  N  Contains  OandGL 

5 

Given  Planes 

PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     91 

ANGLES  BETWEEN  LINES  AND  PLANES 

85.  Problem  26. — To  measure  the  angle  between  any  line  and 
any  plane. 

First  Method. — The  measure  of  the  angle  between  a  line  and 
a  plane  is  the  angle  between  the  line  and  its  projection  on  that 
plane.  Therefore  one  analysis  suggests  itself;  as  follows: 

Analysis. — 1.  Project  the  line  on  the  plane  by  Problem  25. 

2.  Measure  the  angle  between  the  given  line  and  its  projection 
thus  found,  by  Problem  13. 

Let  the  student  make  the  construction. 

85.  Second  Method. — This  method,  while  not  so  obvious  is 
much  quicker  and  better. 

Analysis. — 1.  From  any  point  in  the  given  line  erect  a 
perpendicular  to  the  given  plane. 

2.  Measure  the  angle  between  the  given  line  and  this  perpen- 
dicular. 

3.  Subtract  this  angle  from  90°,   which  gives  the  required 
angle  between  the  line  and  plane. 

Proof. — By  letting  fall  a  perpendicular  to  the  plane,  a  right- 
angled  triangle  is  formed  between  the  given  line,  its  projection, 
and  the  perpendicular.  The  angles  between  the  line  and  its 
projection,  and  between  the  line  and  the  perpendicular  are  com- 
plementary, and  add  up  to  90°.  Hence  this  method  might  be 
called  the  "complementary  method." 

Construction. — Let  it  be  required  to  measure  the  angle  between 
AB  (Fig.  61)  and  the  plane  T. 

1.  From  B  draw  the  line  bvpv  —  bHpH  perpendicular  to  T. 

2.  Through  the  #-piercing  points  of  AB  and  BP  (OH  and  pH) 
draw  HQ. 

Note. — If  the  F-piercing  points  are  more  convenient,  they  may  be  used  in 
preference. 

3.  Revolve  B  into  H  about  HQ  to  bi. 

4.  Lay  off  a  perpendicular  bimi  to  pHbi. 

Conclusion. — The  required  angle  will  be  6,  the  complement  of 
the  angle  between  the  line  and  the  projector. 


92 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


86.  EXERCISES 

Measure    the    angles  that  the    following   lines   make  with   the   planes 
given. 
<H§^  A(l  +  2  -  3)  B(3  +  1  -  1)  and  T(2  +  3)  2(5  -  2). 

486.  C(2  +  3  -  K)  D(4  +  1  -  1)  and  8(5  +  3)  5(1  -  2). 

487.  E(2  +  1  -  2  )F(4  +  1-2)  and  R(l  +  3)  4(1  -  2). 

488.  G(2  +  1  -  K)  K(2  +  2  -  3)  and  R(Ex.  487). 

489.  L(4  +  3  -  K)  M(l  +1-2)  and  R(Ex.  487). 

490.  N(2  -  IK  +  1)  P(4  -  IK  +  1)  and  Q(5  -  2)  2(5  -  IK). 

491.  O(l  +  2  -  1)  K(3  +  1-1)  and  Q(Ex.  490). 

492.  A (2  +  2  -  1)  B(l  +  K  -  3)  and  U(  +  2) « (  -  1). 

493.  C(2  +  3  -  2)  D(2  +  1,  0)  and  U(Ex.  492). 


b, 


494.  E(l  -  3  +  K)  F(2  -  2  +  2)  and  W(  +  1)  «»  (  -  2). 

495.  G(3  +  1  -  1)  K(4  +  K  -  IK)  and  X(l  +  1)  3(5  -  1). 

496.  L(l  +  2  - 

497.  M(2  -1 

498.  O(2  +  2  - 

499.  A(l  +  K 
(1  -  IK)- 

500.  C(2  +  2  -  2)  D(2  +  1  -  K)  and  Z(Ex.  499). 

501.  E(2  +  1  -  3)  F(3  +  3  -  2)  and  Q(l  +  3)  2(5  -  1). 

502.  A(l  +  2  -  2K)  B(2  +  1  -  1)  with  H,  V,  and  P. 

503.  Any  of  the  foregoing  lines  with  H,  V,  and  P. 


K)  E(4  +  2  -  K)  and  X(Ex.  495). 
2)  N(3  -  2  -  K)  and  Y(l  +  3)  5(1  +  1). 

)  P(2  +  K  -  3)  and  Y(Ex.  497). 
K)    B(2  +  3  -  3)     and     Z(l  +3)     (5  +  2) 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     93 

504.  The  angle  of  the  coal  shaft  (Ex.  460)  and  H,  V,  and  P. 

505.  The  angle  of  the  guy  wire  in  Ex.  461,  and  H,  V,  and  P. 

506.  The  angle  of  the  pulley  shaft  (Ex.  462)  and  H,  V,  and  P. 

507.  The  angle  of  the  steam  pipe  (Ex.  256)  and  H,  V,  and  P. 

508.  The  angle  of  the  hip-rafter  AB  (Fig.  34)  with  H,  V,  and  P. 


m  n 


7a  a 


10 


v .  y 

g,k 


Given   Lines 


^ 


I 


HX 


VY 
HY 


z 


10 


12 


lanN  Contains  OandSL 


Given  Planes 
Graphic  Layout  No.  36. — Measure  the  angle  between  line 


and  plane 


94 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


509.  Draw  a  line  from  M(4  +  3  -  3)  making  an  angle  of  60°  with  T(l  +  2) 
5(1  -  3)  intersecting  the  line  A(3  +  2,  0)  B(l,  0  -  3).  What  angle 
does  it  make  with  AB? 

510.  Draw  the  projections  of  the  locus  of  all  lines  from  O(3  +  2  —  3)  making 
67^°  with  S(5  +  2)  2(5  -  1^)- 

511.  Draw  a  line  from  N(2  +2-2)  making  45°  with  R(  +  2) «  (  -  1) 
intersecting  C(2  +  2,  0)  D(4,  0  -  1). 

512.  Draw  a  line  from  P(2  +  3  -  2)  at  60°  from  U(l  +  IK)  3(1  - 
intersecting  E(l  +  IK,  0)  F(l,  0  -  IK). 


HY 


!/ 


Graphic  Layout  No.  37. — Measure  the  angle  between  the  line  and  plane  in 
Ex. . 

513.  Draw  the  projections  of  the  locus  of  all  lines  through  M(3  +2—  1) 
making  60°  with  W(  -  1)  »  (  +  2). 

514.  Draw  a  line  from  P(3,  0,  0)  making  45°  with  T(l  +  2)  (5  +  1)  (1  -  3) 
(5  -  IK),  intersecting  Q(2  +  1%,  0)  K(3,  0  -  2^). 

515.  Draw  the  locus  of  all  lines  through  N(3  +  2,  0)  making  67^°  with  the 
plane  through  GL  and  F(2  +  1  -  2). 


r.^\^ 


^->"  X 


FIG.  62. 


PARALLEL  PLANE  PROBLEMS 

87.  Problem  27. — To  pass  a  plane  through  a  given  point  parallel 
to  a  given  plane. 

Principle. — Corresponding  traces  of  parallel  planes  are  parallel. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     95 

Construction. — Let  O  (Fig.  62)  be  the  given  point.     Required 
to  pass  a  plane  through  O  parallel  to  T. 

1.  Locate  QT. 

2.  Locate  oq. 

3.  Draw  QS  through  OQ  parallel  to  QT. 

4.  Where  QS  intersects  HQ,  draw  HS  parallel  to  HT.     Then 
draw  VS  parallel  to  VT. 

Conclusions. — The  plane  S  contains  O  and  is  parallel  to  T. 


88. 


EXERCISES 


HU 


Z- 


VU 


v 


VK 
HK 


VM 
HM 


10 


VN 


12 


vo 


HO 


HN 


Graphic  Layout  No.  38. — Assume  a  point  in  space,  and  pass  a  plane 
through  it  ||  to  plane .     What  is  the  distance  between  the  planes? 

516)  Through  A(l  +  2  -  J^)  pass  a  plane  ||toR(3  +  3)3(5  -  3).      What  is 
the  distance  between  planes? 

517.  Through  B(3  +2-2)  pass  a  plane  ||  to  Q(l  +  1)  3(1  -  3).     What  is 
the  distance  between  planes? 

518.  Through   C(l  -  1  +  1^)  pass  a   plane   ||  to  S(5  +2)   1(5  -  1^)- 
What  is  the  distance  between  planes? 

519.  Through  D(2  +  1  -  M)  pass  a  plane  ||  to  T(  +  3)  » (  -  2).     What  is 
the  distance  between  planes? 

520.  Through  E(l  +3  —  1)  pass  a  plane  ||  to  the  plane  containing  GL  and 
P(3  +  1  —  2).     What  is  the  distance  between  planes? 


96  PRACTICAL  DESCRIPTIVE  GEOMETRY 

521.  Through  F(4  +  1  -  K)  pass  a  plane  ||  to  U(l  +  2)  3(5  -  2).     What  is 
the  distance  between  planes? 

522.  Through   G(2  +  1  -  2)  pass   a   plane   ||  to   W(l  +  2>£)   4(1  +  1) 
What  is  the  distance  between  planes? 

523.  Through  K(3  +  Y*  -  1)  pass  a  plane  ||  to  X(l  +  3)  (5  -  2)  (1  -  3) 
(5  —  2).     What  is  the  distance  between  planes? 

89.  Problem  28. — To  pass  a  plane  parallel  to  a  given  plane 
at  a  given  distance  from  it. 

Construction. — Let  T  (Fig.  63)  be  the  given  plane.  Required 
to  pass  a  plane  1  in.  from  T  and  parallel  to  it. 

It  is  obvious  that  two  such  planes  can  always  be  passed. 


1.  Obtain  QT. 

2.  Draw  QS  [|  to  and  1  in.  from  QT. 

3.  Where  QS  crosses  HQ  draw  HS  ||  to  HT. 

4.  Draw  VS  ||  to  HT. 

Conclusion. — The  plane  S  is  parallel  to  T  and  1  in.  from  it. 

EXERCISES 

90.  Draw  one  or  two  planes  (according  to  directions  from  instructor) 

J.rallel  to  and  1  in.  from  the  following  planes. 
!4^.'T(1  +  3)  1(4  -  2). 

525.  S(2  +  3)  5(2  -  2). 

526.  Q(4  -  2)  1(5  +3). 

527.  R(  +2)»(  -  1). 

528.  U(  +  !)»(  +2). 

529.  ~W(  -  !)»(  - 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     97 


530.  X(  -2)co(  -  1). 

531.  Y(5  +  2)  3(1  -  2). 

532.  Z(5  +  3)  1(5  +  1). 

533.  Q(5  +  2)  (1+  1)  (5  -  3)  (1  - 

534.  R(5  +  H)  (1  +  IK)  (5  +  2)  (1  +  3). 

535.  Scale  %-in.  =  1ft.    A  saw-tooth  roof  is  being  erected  for  a  greenhouse. 
The  planes  are  all  parallel  to  T(42  ft.  +  8  ft.)  34  ft.  (42  ft.  -  4  ft.) 
Draw  as  many  of  the  roof  planes  as  possible  in  your  space  at  a  distance 
of  9  ft.  between  each. 

536.  The  alternate  planes  in  the  foregoing  exercise  are  perpendicular  to  T. 
Let  the  ridge  line  be  7  ft.  6  in.  above  and  parallel  to  H.     Draw  as  much 
of  the  outline  of  the  greenhouse  as  is  possible  in  your  space. 


1             s, 

&/ 

/ 

'     V 

3  HU  

'   */ 

y 

Sr^. 

~z. 

~"vu~ 

/ 

5-*^& 

6\ 

V 

7  ^. 

^vz^X. 

8 

^  — 

»>. 

VK 

HK 

9        VM 
HM 

10 
VN  

'4 

ie        
-  —  vo  
"HO 

HN 

%\ 

Graphic  Layout 

...                     .  .       ,     . 

(1     1  !/•    11Z    13, 

i,  2)  in. 

DIHEDRAL  ANGLES 


91.  Problem  29. — To  measure  the  dihedral  angle  between 
two  planes. 

The  angle  between  two  planes  is  considered  to  be  equal  to 
that  angle  included  between  the  perpendiculars  to  the  line  of 
intersection,  that  lie  in  each  plane.  It  is  evident,  then,  J,hat 
the  plane  of  the  measuring  angle  is  perpendicular  to  the  line 

7 


98 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


of  intersection,  and  therefore  to  both  of  the  given  planes.     Two 
methods  of  finding  this  angle  are  given  herewith. 

First   Method. — Analysis. — 1.  Find   the   intersection   of   the 
given  planes. 

2.  At  any  assumed  point  on  the  line  of  intersection  pass  a 
plane  perpendicular  to  it. 

Note. — Only  one  trace  of  this  auxiliary  plane  is  usually  necessary. 

3.  Measure  the  angle  between  the  intersections  of  the  auxiliary 
plane  with  the  two  given  planes. 

Construction. — Let  S  and  T  (Fig.  64),  be  the  two  planes. 


FIG.  64. 

1.  Obtain  their  line  of  intersection,  AB. 

2.  Through  0,  assumed  on  AB,  draw  OP,  perpendicular  to 
AB  and  parallel  to  V. 

3.  Obtain  P,  the  H-piercing  point  of  OP. 

4.  Draw  HQ  through  pH,  perpendicular  to  aHbH. 

5.  Connect  OH  with  x  and  y  (intersections  of  HQ  with  HT  and 
HS). 

6.  Revolve  0  to  OL 

Conclusion. — X0iy(0)  is  the  required  angle. 
Second  Method. — Analysis. — 1.  From  any  point  let  fall  two 
perpendiculars,  one  to  each  of  the  two  given  planes. 

2r  Measure  the  angle  included  between  these  perpendiculars. 
This  will  be  the  correct  measure  of  the  required  dihedral  angle. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     99 

Proof. — The  two  perpendiculars  from  the  point  to  the  planes 
determine  a  plane  perpendicular  to  both  planes.  Fig.  65  shows 
a  view  of  two  intersecting  planes  S  and  T.  O  is  taken  anywhere, 
and  perpendiculars  OM  and  ON  are  let  fall  to  T  and  S  respectively. 
The  plane  of  MON  cuts  lines  MP  and  NP  from  T  and  S,  and  the 
four  lines,  OM,  ON,  MP  and  NP  form  a  quadrilateral,  the  sum 
of  whose  angles  is  360°.  Prove  this.  The  angles  at  M  and  N 
are  right  angles  by  construction,  and  therefore  add  up  to  180°. 
This  leaves  180°  as  the  sum  of  angles  <f>  and  6,  at  0  and  P  re- 
spectively. The  angle  $  is  therefore  the  supplement  of  0,  which 
is  the  angle  between  T  and  S.  But  the  measure  of  any  angle, 
plane  or  dihedral,  may  be  either  the  acute  angle  or  its  supplement, 
so  the  angle  between  the  planes  S  and  T  may  be  either  6  or  </>, 


FIG.  65. 

and  as  <f>  is  the  angle  between  the  perpendiculars,  it  is  also  the 
measure  of  the  given  dihedral  angle. 

Note. — Although  the  angle  between  two  lines,  or  two  planes,  may  be 
either  the  acute  or  the  obtuse  angle,  as  stated  in  the  foregoing,  the  acute 
angle  is  usually  taken  as  the  true  measure. 

92.  EXERCISES 

Dimensions  of  planes  for  following  exercises: 
8(5  +  2)  1(5  -  3). 

T(+2)o>(  -1). 

R(   _l)oo(_l^). 

Q(5  +  3)  5(5  -  2). 
X(l  +  2)  3(5  -  2). 
W(5  -  2)  4(1  -  1). 
Y(l  -  3)  4(1  -  1). 
Z(5  -  3)  (1  -  2)  (5  +  IK)  (1  +  1). 
P(3  +  3)  3(3  -  3). 
M(l  -  3)  (5  -  IK)  (1  -  2)  (5  -  1). 
U  containing  A (2  +  1  -  Ifi)  and  GL- 
537.  Measure  the  dihedral  angle  of  the  planes and . 


100 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


538.  Measure  the  angles  that  the  plane makes  with  H,  V,  and  P. 

539.  Measure  the  hip  roof  angle  between  the  roofs  C  and  D  in  Fig.  34,  Art.  45. 

540.  A  room  is  18  ft.  long,  10  ft.  wide,  and  10  ft.  high.     The  center  line  of  a 
chute  runs  from  a  point  in  the  ceiling  2  ft.  from  the  right  end  wall 
and  1^  ft.  from  the  rear,  to  a  point  in  the  floor,  5  ft.  from  the  left  end 
wall  and  2  ft.  from  the  front.     The  chute  is  2  ft.  square,  with  the 
diagonal  parallel  to  the  floor. 

(a)  Find  the  floor  opening  for  the  chute. 

(6)   Find  the  angles  of  its  faces  with  the  floor.     Scale  1  in.  =3  ft. 


3      HR 
VR 


vs 


10 


vtt 


Graphic  Layout  No.  40. — Exercises  537  and  538  apply  to  either  graphic 
or  dimensioned  layout. 

541.  A  hexagonal  roof  tower,  12-ft.  base  diameter,  16-ft.  altitude,  has 
specially  rolled  angle  irons  for  its  frame.  What  is  the  angle  at  which 
they  must  be  rolled? 

93.  Problem  30.— To  find  the  traces  of  a  plane  that  shall 
make  a  given  angle  with  a  given  plane  and  contain  a  given  line 
of  that  plane. 

Analysis. — Assume  a  point  in  the  given  line,  and  pass  a  plane 
through  the  point  perpendicular  to  the  line. 

2.  Revolve  the  line  cut  by  this  auxiliary  plane  from  the  given 
plane  about  the  trace  of  the  auxiliary  plane  into  H  or  V,  as  the 
case  may  require. 

3.  From  the  revolved  position  of  the  assumed  point  on  the 
given  line  lay  off  a  line,  making  the  given  angle  with  the  revolved 
line. 

4.  Where  this  new  line  crosses  the  trace  of  the  auxiliary 
plane  will  be  a  point  on  the  trace  of  the  required  plane. 

5.  Since  the  required  plane  must  contain  the  given  line,  its 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     101 

traces  will  pass  through  the  H-  and  F-piercing  points  of  the 
given  line. 

6.  Draw  the  traces  of  the  required  plane  through  the  point  thus 
found  and  the  piercing  points  of  the  given  line. 

Construction. — Let  T  (Fig.  64)  be  the  given  plane,  AB  the 
given  line,  and  6  the  given  angle. 


FIG.  64. 


1.  Assume  0,  any  point  in  AB. 

2.  Through  0  pass  a  plane  perpendicular  to  AB.     HQ  is  the 
trace  of  this  plane. 

3.  Revolve  0  into  H  about  HQ  to  OL 

4.  Lay  off  Oiy  at  angle  0  with  oix. 

5.  Through  bH  and  y  draw  HS. 

6.  Draw  VS  through  av. 

Conclusion. — S  is  the  plane  containing  AB  making  the  angle 
8  with  T. 

Note. — There  can  be  two  planes  passed  through  AB  making  0  with  T, 
except  when  8  =  90°. 

Supplementary  Problem. — Through  a  given  line  to  pass  a 
plane  perpendicular  to  a  given  plane. 

Analysis. — 1.  Through  any  point  of  the  given  line  draw  a 
line  perpendicular  to  the  given  plane. 

2.  Pass  a  plane  through  the  two  lines. 


102 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Conclusion. — This  plane  will  be .  perpendicular  to  the  given 
plane.     Why? 
Let  the  student  make  the  construction. 

Note. — If  the  given  line  is  perpendicular  to  the  given  plane,  the  number  of 
solutions  is  infinite. 

EXERCISES 


1 

\ 

V 

3 

D. 

4 

5            * 

A 

n 

i 

ft. 

x 

J 

la: 

r 

6> 

7_J!T  

8 

vs" 

9 

10     VW 
HW 

\ 

VT 

""HS" 

HU 
VU 

"     V 

^HV— 

13   f      S 

&q^ 

14 

15 

'*\*^* 

^JJ.X.-  — 

^ffc: 

^ 

the  conditions  of  Ex.  542. 

(6)  Plane  --  .     Draw  traces  of  a  plane  under  the  conditions  of  Ex.  543. 
(c)    Plane  --  .     Draw  traces  of  a  plane  under  the  conditions  of  Ex.  544. 

94.  Use  the  following  planes  in  Ex.  542-544,  and  546. 

M(l  +2)  5(1  -  3). 

N(5  -1)2(5  +2^). 

0(1  +  3)  (5  +  \yz]  (1  +  2)  (5  +  1). 

P(3  +  3)  3(3  -  3). 

Q(2  -  3)  2(5  +  2). 

R(l  +  1)  4K(4M  +  3). 

S(5  +  2)  3(1  -  2). 

T(+2)o>(  -1). 


Y(l  +  2)  (5  +  1)  (1  -  2y2)  (5  - 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     103 


Z(5  +  1)  1(5  • 
_^K(1  +3)2(5  -  1).  tf  v 

542.  Assume  a  line  1  in.  from  and  parallel  to (H  or  V)  in  the  plane  •**  — . 

Draw  the  traces  of  plane  that  shall  contain  this  line  and  make  a  dihe- 
dral angle  of °  with  the  given  plane. 

543.  Plane  .     Assume  a  line  running  obliquely  to  H,  V,  and  P  in  the 

plane,  and  draw  the  traces  of  a  plane,  which  contains  the  line,  and 
makes °  with  the  given  plane. 

544.  Plane .     Assume  a  profile  line  in  the  plane,  and  draw  the  traces  of  a 

plane,  which  contains  the  line,  and  makes °  with  the  given  plane. 


M 


H 


'H 


<r  i 


10 


Graphic  Layout  No.  42. — Draw  the  traces  of  two  planes  containing  line 
,  and  making  a  dihedral  angle  of °  (15,  30,  45,  60,  75,  90). 

545.  The  dihedral  angle  between  two  adjacent  faces  of  a  right  pyramid,  base 

a2-in.  equilateral  triangle,  is °  (95,  100,  105,  110,  120,  135,  150). 

Draw  the  projections  of  the  pyramid. 

546.  Assume  a  line  in  space,  and  pass  a  plane  through  the  line,  perpendicular 
to  given  plane .     (Use  either  dimensioned  or  graphic  layout). 

PROBLEMS  UTILIZING  THE  RIGHT  CIRCULAR  CONE 

95.  Properties  of  the  Right  Circular  Cone. 

Three  properties  of  the  cone  are  particularly  useful  in  the  solu- 
tion of  certain  problems. 

1.  The  elements  of  the  cone  all  make  the  same  angle  with  the 
plane  of  the  base,  and  they  are  all  of  the  same  length,  because 
one  way  of  generating  a  cone  is  to  revolve  a  right-angled  triangle 
about  one  of  its  perpendicular  sides. 

2.  All  planes  tangent  to  a  cone  make  the  same  angle  with  the 
plane  of  the  base,  because  a  tangent  plane  is  perpendicular  to 
the  generating  triangle  of  the  cone,  and  contains  its  hypothenuse. 


104 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


3.  Every  element  of  the  cone  and  every  plane  tangent  to  the 
cone  pass  through  the  apex. 

With  these  properties  understood,  it  is  a  simple  matter  to  solve 
the  six  following  problems: 

96.  Problem  31. — To  draw  a  line  through  a  point  so  as  to 
make  given  angles  with  H  and  V. 

Limits. — The  sum  of  these  angles  cannot  be  more  than  90°. 
If  a  line  be  parallel  to  GL,  the  sum  is  0°,  and  if  it  lie  in  a  profile 

plane,  the  sum  is  90°.  For 
any  other  position,  the  sum 
will  be  less  than  90°. 

Construction. — Let  it  be 
required  to  draw  a  line 
through  A,  Fig.  66,  that  shall 
make  45°  with  H,  and  30° 
with  V. 

1.  Make  A  the  apex  of  a 
right    circular    cone,    whose 
base  is  in  H,  and  whose  ele- 
ments make  45°  with  H. 

All  the  elements  make  45° 
with  H,  and  all  pass  through 
A.  Therefore,  it  remains  to 
be  found  which  are  the  four 
possible  elements  that  make 
30°  with  V. 

2.  From    A    draw   a  line 
avbv,  aHbi  parallel  to  H,  making  30°  with  V,  and  equal  in  length 
to  the  elements  of  the  cone. 

3.  Let  this  line  be  an  element  of  a  second  cone  with  its  apex  at 
A,  and  base  parallel  to  V.  .  Draw  the  projections  of  the  cone 
(whole  or  in  part). 

4.  Where  the  bases  of  the  cones  intersect  (B  or  E),  the  common 
element  can  be  drawn. 

Conclusion. — The  line  AB  (as  well  as  the  lines  AC,  AD,  and 
AE)  is  inclined  45°  to  H,  and  30°  to  V. 

Let'  the  student  give  the  reason  for  this  and  supply  a  proof  and 
a  means  of  checking  its  correctness. 


TIG.  60. 


97. 


EXERCISES 


547.  Through  M(3  +  \Y2  —  2)  draw  four  lines,  each  making  60°  with  H  and 
15°  with  V. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     105 

548.  Through  O(2  +  2  -  1)  draw  four  lines,  each  inclined  30°  to  H  and  V. 

549.  Draw  a  line  4  in.  long,  terminating  in  H  and  V,  inclined  20°  to  H  and 
40°  to  V. 

550.  Draw  a  line  through  K(4  +  1  -  1)  making  35°  with  H  and  55°  with  V. 

551.  Draw  a  line  through  P(3^,  0,  0)  making  20°  with  H  and  50°  with  7. 

98.  Problem  32. — To  measure  the  angle  an  oblique  plane  makes 
with  H  or  V. 

Construction. — To  measure  the  angle  between  T  and  H  (Fig. 
67). 

1.  With  any  assumed  point  (O)  on  VT  as  an  apex,  draw  a  cone 
with  its  base  tangent  to  HT. 


FIG.  67. 


Conclusion. — The  angle  6,  the  angle  between  the  elements 
and  the  base  plane,  is  the  measure  of  the  inclination  of  T  to  the 
horizontal. 

Notes. — 1.  The  angle  <f>  shown  in  Fig.  67  is  the  measure  of  the  angle 
between  T  and  V. 

2.  This  problem  was  also  solved  incidentally  in  Problem  2. 


99. 


EXERCISES 


Measure  the  angles  that  the  following  planes  make  with  H  and  V. 

552.  Q(l  +  2)  5(1  -  3). 

553.  R(5  -  1)  2(5  +  2Y2). 

554.  0(1  +  3)  (5  +  IY2)  (1  -  2)  (5  -  1). 

555.  S(5  +  2)  3(1  -  2). 

556.  T(  +  2)  *  (  -  1). 

557.  U(  _  l^)«>  (+2) 

558.  W(+2)co(  +  ^) 

559.  X(  -  1H)«(  -  1). 

560.  Y(l  +  2)  (5  +  1)  (1  +  2^)  (5  +  1H). 

561.  Z(5  +  1)  2(5  +  3). 


106 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


562.  K(l  +  3)  2(5  -  1). 

563.  N  through  GL  and  E(2  +  2  -  H)- 

564.  A  tetrahedron  is  limited  by  the  points  A(2,  0  -  1),  B(3^,  0 

C(4,  0  -  3),  and    D(3  +  1^  ~  2^).     Measure  the   dihedral  angle 
between  face and  face . 


565.  Tetrahedron  (Ex.  564). 
with  H  and  V. 


Measure  the  angles  of  the  three  oblique  faces 


\l 

t' 

3 

^ 
t 

4 

~"vw~ 

5 
HX 

6  */ 

-\* 

s<& 

X 

^2/^ 

~w"~ 

JOL_ 

*/ 

7 

\ 

8  V 

9         ^ 

-"HZ~ 

10 

^M 

—  -^ 

X 

2-r  i  

_!!m 

\\ 

"i 

^£^ 

S& 

a 

°\ 

Graphic  Layout  No.  42. — Measure  the  angles  between  given  plane 

and  H  and  V. 

100.  Supplementary  Problem  33 (a). — Given  one  trace  of  a 
plane  and  the  angle  between  the  plane  and  the  plane  of  projec- 
tion, to  find  the  other  trace. 

Analysis. — 1.  Construct  a  cone,  whose  elements  make  the 
given  angle  with  its  base,  tangent  to  the  given  trace,  with  the 
center  of  its  base  in  GL. 

2.  Draw  the  other  trace  through  the  apex  of  the  cone  and  the 
intersection  of  the  given  trace  with  GL. 

Notes. — 1.  If  the  given  trace  does  not  intersect  GL  within  bounds,  two 
cones  can  be  constructed,  and  the  trace  drawn  through  the  two  vertices. 

2.  If  the  given  trace  is  parallel  to  GL,  the  other  trace  will  also  be  parallel, 
and  the  quickest  solution  is  effected  through  the  profile  plane. 

3.  By  inverting  the  cone,  another  plane  could  be  found  that  would  fulfil 
the  conditions. 

101.  Supplementary  Problem  33(b). — Given  the  H-trace  of  a 
plane  and  its  angle  with  V,  to  find  the  V-trace. 

Analysis. — 1.  From  any  point  in  the  given  trace  as  an  apex, 
construct  a  cone  of  the  required  elemental  angle,  having  its  base 
in  V. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     107 

2.  From  the  intersection  of  the  given  trace  with  GL  draw  the 
F-trace  tangent  to  the  base  of  the  auxiliary  cone. 

Note. — See  notes  following  33 (a). 

102.  EXERCISES 


\ 


\ 


Graphic  Layout  No.  43. — H-  or   F-traces  of  planes  to  be  used  with  Ex. 
566-579. 

566.  Given //-traces:  S5(l  -  2),  T4(2  -  3),  U  °°(  _  1),  W(l  -  3)  (5  -  2), 
X5(l  +  IK),  Y  oo  (  +  1^),  Z(l  +  2)  (5  +  1). 

Draw  the  F-trace  of  plane ,  when  the  plane  makes  • °  with  H. 

567.  Draw  the  F-trace  of  plane  (Ex.  566),  when  the  plane  makes 

°  with  F. 

568.  The  //-trace 
to  H  at  


-  (Ex.  566)  is  the  //-trace  of  two  planes,  each  inclined 
Draw  their   F-traces.     Measure  the  angle  between 
the  two  planes. 

569.  Draw  the  traces  of  the  bisecting  plane  of  the  dihedral  angle  formed  by 
the  two  required  planes  in  Ex.  568. 

570.  Given  trace (Ex.  566).     This  is  the  H  -trace  of  two  planes  mak- 
ing   °  with  F.     Draw  their  F-traces. 

571.  Measure  the  dihedral  angle  between  the  two  required  planes  in  Ex.  570. 

572.  Draw  the  traces  of  the  bisecting  plane  of  the  dihedral  angle  between  the 
required  planes  in  Ex.  570. 

573.  Given  F-traces:  X(3  +  3)2,  Y(5  +  1)1,  Z(  +  2)  <*>,  M(5  +  2)(1  +  1), 
S(l  -  1^)5,  T(2  -  3)3,  U(  -  1)  co,  W(l  -  2K)  (5  -  1).     Draw  the 
//-trace  of ,  when  the  plane  makes °  with  F. 

574.  Draw  the  //-trace  of  -    —  (Ex.  573),  when  the  plane  makes  — ; — ° 
with  H.     What  is  its  angle  with  F? 

575.  The  F-trace  (Ex.  573)  is  the  trace  of  the  two  planes  making 

°  with  F.     Draw  their  //-traces. 

576.  The  F-trace  (Ex.  573)  is  the  trace  of  a  plane  making  ° 

with  H  and  another  plane  making °.     Draw  their  //-traces. 


108 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


577.  Measure  the  dihedral  angle  between  the  two  planes  required  in  Ex.  575. 

578.  Measure  the   dihedral   angle  between  the  two  planes  required  in 
Ex.  576. 

579.  Draw  the  traces  of  the  plane  bisecting  the  dihedral  angle  between  the 
two  planes  required  in  Ex.  575  or  Ex.  576. 

580.  Through  a  point  A  (3,  0  —  1)  in  a  vein  of  ore  a  horizontal  line  on  the 
surface  runs  north,  37°  west.     The  "dip"  (inclination  of  the  ore  to  the 
horizontal)  is °.     Find  the  F-trace  of  the  plane  of  the  ore. 

581.  The  lower  edge  of  a  rectangular  roof  is  30  ft.  long  and  inclined  40°  to 
the  F-plane.     The  roof  is  inclined  60°  to  the  horizontal  and  the  ridge  is 
8  ft.  higher  than  the  lower  edge.     Draw  the  H-  and  F-projections  of 
the  roof,  and  insert  three  windows,  each  2  ft.  6  in.  X  6  ft.  3  in.,  sym- 
metrically placed.     Draw  to  a  scale  %  in.  =  1  ft. 

582.  Given  the  line as  the trace  of  a  plane  inclined °  to . 


583. 


Draw  the  other  trace. 
The  line  is  the 


What  angle  does  it  make  with 


trace  of  two  planes  making 


with 


.     Draw  their  traces. 

584.  The  line  —  —  is  the trace  of  two  planes,  one  making °  with 

— ,  and  the  other °  with .     Draw  their  traces. 

585.  Measure  the  dihedral  angle  between  the  required  planes  in  Ex.  583  or 
584. 

586.  Draw  the  traces  of  the  bisector  of  the  dihedral  angle  of  the  required 
planes  in  Ex.  583  or  584. 


FIG.  68. 


103.  Problem  34. — To  pass  a  plane  through  a  given  line,  so  as 
to  make  any  required  angle  with  a  plane  of  projection. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     109 

Construction. — Let  AB  (Fig.  68)  be  the  given  line,  and  let  it 
be  required  to  pass  a  plane  through  it,  which  is  inclined  60°  to 
the  horizontal. 

1.  Locate  the  H-  and  F-piercing  points  of  AB. 

2.  Using  any  point  (as  B)  as  the  apex  of  a  60°  cone  on  H,  draw 
the  projections  of  the  cone,  as  shown. 

3.  Draw  HT  through  E  tangent  to  the  base  of  the  cone. 

4.  Draw  VT  from  T  through  O. 

Note. — There  are  two  possible  solutions,  except  when  the  given  angle 
is  90°. 

Limitation. — If  the  inclination  of  the  given  line  exceeds  the 
angle  of  the  required  plane  with  the  given  plane  of  projection, 
the  problem  is  impossible.  Why? 

104.  EXERCISES 

587.  Given  lines:  A(2  +  3  -  2)  B(4  +  1  -  H)- 

C(2  +  1  -  3)  D(4  +  1  -  y2}. 
E(l  +  2K  -  H)  F(4  +  IK  -  1H). 
G(3  +  1  -  K)  K(3  +  2  -  3). 
L(3  +  2  -  Y2)  M(3  +  H  -  3). 

N(2  +  IK  -  i)  0(4  +  iy2  -  i). 

P(2  -  2  +  %)  X(4,  0  +  %). 

Y(3  -  3  +  1)  Z(5,  0  +  2). 

M(3  -  1  +  2)  P(3  +  2  +  M). 

A(2  +  2  +  2)  F(4  +  2  +  2). 

Pass  a  plane  through  line making °  with .     Measure  its 

angle  with  the  other  planes  of  projection. 

588.  Draw  the  traces  of  two  planes  containing  the  line  -    -  (Ex.  587), 

each  making °  with .     Measure  their  angles  with  the  other 

planes  of  projection. 

589.  Draw  the  traces  of   two  planes  containing   line (Ex.  587),  one 

inclined °  to  ,  and  the  other  °  to .     Measure  their 

angles  with  the  other  planes  of  projection. 

590.  Measure  the  dihedral  angle  between  the  planes  required  in  Ex.  588 
or  589. 

591.  The  line  E(2  +  3  -  1)  F(4,  0  -  2)  is  the  hip  rafter  of  a  roof  each  of 

whose  sides  is  an  isosceles  triangle,  and  is  inclined °  to  H.     Draw 

the   projections   of   the  roofs.     Measure  the  hip  angle.     Draw  the 
projections  of  a  window  in  the  center  of  each. 

592.  Through  given  line  pass  a  plane  inclined °  to .     Meas- 
ure its  angles  with  the  other  planes  of  projection. 

593.  Through  given  line pass  two  planes  inclined  —    -°  to .     Meas- 
ure their  angles  with  the  other  planes  of  projection. 

594.  Through  given  line pass  a  plane  inclined °  to ,  and  an- 
other   °  to .     Measure  their  angles  with  the  other  planes  of 

projection. 


110 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


595.  Measure  the  dihedral  angle  between  the  two  planes  required  in  Ex.  593 
or  Ex.  594. 

596.  Use  one  of  the  given  lines  as  a  roof  ridge  or  rafter  and  draw  the  roof  as 
directed  by  the  instructor,  triangular,  rectangular,  etc.,  at  any  inclina- 
tion given,  and  with  or  without  windows. 


a     .a" 

bH 


a 
bH 


I 


a 
bv 


a 


r^ 

bv 


10 


12 


Graphic  Layout  No.  44. — To  be  used  with  Ex.  587-596. 

105.  Problem  35. — To  find  the  traces  of  a  plane  which  shall 
make  any  required  angles  with  H  and  V. 

Limitations. — The  sum  of  these  angles  must  be  between  the 
limits  of  90°  and  180°.  A  plane  parallel  to  H  or  V  is  perpen- 
dicular to  the  other.  A  profile  plane  is  90°  to  each.  Oblique 
planes  are  between  these  limits. 

Construction. — Let  it  be  required  to  draw  the  traces  of  a  plane 
inclined  45°  to  V  and  60°  to  H. 

1.  Draw  the  projections  of  a  sphere  of  any  diameter  with  its 
center  in  GL.     (See  Fig.  69.) 

2.  Draw  a  45°  cone  tangent  to  the  sphere,  having  its  base  on 
V  and  apex  in  H. 

3.  Draw  a  60°  cone  tangent  to  the  sphere,  having  its  base  on  H 
and  apex  in  V. 

4.  Draw  VT  through  the  apex  in  V,  tangent  to  the  base  of  the 
other  cone,  and  HT  through  the  apex  in  H,  tangent  to  the  base 
that  is  in  H. 

Conclusion. — This  plane  tangent  to  both  cones  thus  fulfils  the 
conditions  of  the  problem 

Notes. — 1.  Eight  planes  fulfilling  these  conditions  can  be  drawn,  all 
tangent  to  the  sphere. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     111 

2.  Any  plane  parallel  to  GL  has  the  sum  of  its  angles  with  H  and  V  equal  to 
90°,  hence,  if  the  sum  of  the  given  angles  is  equal  to  90°,  the  problem  can  be 
easiest  solved  by  the  use  of  the  profile  plane. 

3.  If  it  be  required  to  pass  such  a  plane  through  a  given  point,  an  auxiliary 
plane  may  be  drawn  anywhere,  as  in  the  foregoing,  filling  the  conditions, 
and  then  a  plane  may  be  passed  through  the  point  parallel  to  the  auxiliary- 
plane.     The  plane  S,  Fig.  69,  is  passed  through  the  point  O,  parallel  to  T 
already  found.     The  plane  S  makes  the  same  angles  with  H  and  V  that  T 
does. 

Second  Method. — (From  Church  and  Bartlett.) — "If  a  line 
is  perpendicular  to  a  plane,  the  angles  between  the  line  and  the 
planes  of  projection  will  be  complements  of  the  angles  between 
the  plane  and  the  planes  of  projection.  Hence  if  we  construct 


/'          ft                         H 
I    /,'            a 

rx  \   r    <> 

\  \/ 

\  A     / 

ov 

x\u 

1  1  \ 

----  -V 


FIG.  69. 


a  line,  making  with  H  and  V  the  complements  of  the  required 
angles,  a  perpendicular  plane  through  this  point  will  make  the 
required  angles  with  H  and  V." 


106. 


EXERCISES 


to  V  and  -^  —  °  to  H  through 

-  °  to  V  and  --  °  to  H 
(6)  Assume  the  point  in  II.     (c) 


597.  .1  Draw  the  traces  of  a  plane  inclined  —  —  ° 

the  point  A  (3  +  1  -  2). 

598.  (a)  Draw  the  traces  of  a  plane  inclined 
through  an  assumed  point  in  ///. 
Assume  the  point  in  IV. 

599.  Pass  a  plane  under  the  conditions  in  Ex.  597  or  598,  using  the  second 
method. 

600.  A  line  shaft  making  -  °  with  H  and  -  °  with  V  carries  a  12-in. 
pulley  with  6-in.  face,  with  its  center  at  M(3  ft.  6  in.,  +2  ft.  —  1  ft.  8 
in.).     Draw  the  projections  of  the  pulley  to  a  scale  1  in.  =   1  ft. 


112 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


601.  A  pulley  24  in.  X  8  in.,  whose  center  plane  is  inclined °  to  H  and 

°  to  V,  runs  on  a  shaft  8  ft.  long  extending  from  a  point  in  the 

wall  (F)  to  a  point  in  the  floor  (H).     The  pulley  is  in  the  center  of  the 
shaft.     Draw  its  projections. 

602.  A  man  is  reading  a  newspaper  20  in.  X  30  in.  and  his  eye  at  D(10  ft.  + 
4  ft.  —  6  ft.)  is  2  ft.  distant  from  the  paper.     His  line  of  sight  is 

inclined  °  to  H  and  —  -°  to  V,  and  travels  to  the  geometrical 

center  of  the  sheet.     Draw  the  projections  of  the  paper. 

608.  Assume  a  point  in  the angle,  and  draw  the  traces  of  all  the  planes 

that  can  contain  the  point  and  be  inclined °  to  H  and °  to  V. 

SPECIAL  PROBLEMS 

107.  Problem  36. — To  find  the  shortest  distance  between  two 
lines  not  in  the  same  plane. 


Construction. — Required  the  shortest  distance  between  AB 
and  CD  (Fig.  70). 

Referring  to  Fig.  71,  we  find  that,  when  a  line  is  perpendicular 
to  H  or  V  (as  AB  is  to  H),  the  common  perpendicular  is  immedi- 
ately drawn,  without  further  operations  on  AB  and  CD.  Why? 
Therefore,  this  problem  is  easiest  solved  by  bringing  one  of 
the  lines  into  this  relation. 

Return  to  Fig.  70.     Draw  Q  parallel  to  CD. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     113 

1.  Obtain  the  Q-projections  of  AB  and  CD. 

2.  Project  them  on  X,  which  is  perpendicular  to  CD. 

3.  Draw  the  perpendicular,  y*xx,  from  cxdx  to  axbx.     This  is 
the  shortest  distance. 


FIG.  71. 

108.  Supplementary  Problem. — To  find  the  shortest  line,  or 
"common  perpendicular." 

Construction. — Taking  the  X-projection  of  the  shortest 
distance,  xxyx,  projecting  it  back  to  the  Q-projections,  and  thence 
to  the  H-  and  F-projections,  we  obtain  the  H-  and  F-projections 
of  the  shortest  line. 


Fia.  72. 

109.  Second  Method. 

Analysis. — 1.  Pass  a  plane  through  one  of  the  lines  parallel 
to  the  other. 

2.  Project  the  outside  line  on  the  plane. 

8 


114  PRACTICAL  DESCRIPTIVE  GEOMETRY 

3.  At  the  intersection  of  this  projection  with  the  given  line, 
erect  a  perpendicular  from  the  plane  to  the  outside  line.  This 
line  is  the  common  perpendicular. 

Construction. — Let  AB  and  CD  (Fig.  72)  be  the  given  lines. 

1.  Draw  EF  parallel  to  AB  intersecting  CD. 

2.  Pass  the  plane  Q  through  EF  and  CD. 

3.  'Project  AB  to  ab  on  Q. 

4.  Where  ab  crosses  CD  at  o,  erect  the  common  perpendicular 
oO. 

This  method,  the  standard  for  many  years,  is  still  preferred 
by  many.  Let  the  student  make  the  construction  in  orthographic 
projection. 

110.  EXERCISES 

Draw  the  projections  of  the  common  perpendicular  to  the  following 
lines : 

604.  A(l  +  2  -  2)  B(4,  0-1)  and  C(3,  0-3)  D(4K  +2-1). 

605.  E(4  +  1  -  1)  F(4  +  1  -  3)  and  G(l  -  1  -  1)  K(3  -  3  -  3). 

606.  M(l  +  2  -  1)  N(3  +  2  -  1)  and  O(2  +  3  -  3)  P(3  +  1  -  2). 

607.  A(2  -  2  +  1)  B(3K,  0+3)  and  C(2  -  1J$,  0)  D(4  -  3  +  1). 

608.  E(2  +  3  -  H)  F(2  +  1  -  2^)  and  G(3  +  3  -  3)  K(C,  0,  0). 

609.  M(2  +  3,  0)  N(2,  0-2)  and  O(2>£,  0-3)  P(4,  0  +  2>£). 

610.  X(2  +  3,  0)  Y(3K,  0-2)  and  GL. 

611.  A(l  +  1  -  1)  B(4  +  1  -  3)  and  C(2  +  3  -  1)  D(4,  0  -  1). 

612.  E(l  +  3  -  1)  F(3,  0-1)  and  G(3  +2^-2)  K(4,  0,  0). 

613.  M(l  +  2  -  M)  N(2,  0-2)  and  O(3  +  3  -  3)  P(3M  +  1  -  M)« 

614.  Draw  the  shortest  possible  line  parallel  to (H,  V,  or  P)  intersecting 

both  lines  in  Ex.  -   -  (604-613). 

615.  Given  a  2-in.  cube.     Draw  the  shortest  line  between  the  non-intersect- 
ing diagonals  of  any  two  adjacent  faces. 

616.  Two  wires  run  through  a  building,  A  (12  ft.  +  23  ft.,  0)  B(28  ft.,  0  - 
6  ft.)  and  C(16  ft.  +  12  ft.,    0)  D(24  ft.,  0-16  ft.).     Draw  the 
shortest  connecting  wire.     Scale  %  in.  =  1  ft. 

617.  A  2^-in.  pipe  is  run  from  a  hole  in  a  wall  at  E(13  ft.  +  12  ft.,  0)  at 
45°  to  H  and  45°  to  V.     At  F(8  ft.  +  6  ft.,  0)  a  1-in.  pipe  runs  straight 
out  from  the  wall,  then  bends  perpendicularly  and  runs  downward  to 
the  right  at  40°  to  H .     Make  the  shortest  possible  connection.     How 
long  must  it  be,  allowing  2  in.  at  each  end  for  connections?     Scale 
Kin.  =lft. 

618.  Draw  the  shortest  possible  horizontal  pipe  for  Ex.  617. 

619.  Draw  the  shortest  possible  vertical  pipe  for  Ex.  617. 

620.  Scale  }/±  in.  =  1  ft.    In  a  mill  there  is  a  6-ft.  cubical  bin  on  the  floor  with 
one  edge  at  A(15  ft.,  0,  0)  B(21  ft.,  0,  0).    A  24-in.  square  chute  runs 
its  center  line  from  C(3  ft.  +  9  ft.,  0)  in  the  wall  to  D(16  ft.,  0-10 
ft.)  hi  the  floor.     It  is  desired  to  run  a  9-in.  cylindrical  pipe  from  a  hole 
at  E(18  ft.  +  10  ft.,  0)  to  F(8  ft.,  0-13  ft.).     Can  the  pipe  pass  both 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     115 


the  bin  and  the  chute?     If  so,  how  much  will  there  be  to  spare  in  each 
case?     If  not,  locate  the  nearest  point  to  transfer  F. 


In" 


•  i 


L 


k" 


Graphic  Layout  No.  45. — (Note. — Symbols  may  be  transposed  at  option  of 
instructor.) 


621.  Draw  the  shortest   line   between  the    two  lines  in  Ex. 
out  45. 


of  Lay- 
of 


622.  Draw  the  shortest  horizontal  between  the  two    lines  in   Ex.  - 
Layout  45. 

623.  Draw  the  shortest  vertical  between  the  two  lines  in  Ex.  of  Lay- 
out 45. 

624.  Draw  the  shortest  line  between  GL  and  any  oblique  line  in  Layout  45. 


FIG.  73. — Pulleys  on  non-intersecting  shafts. 

111.  The  Problem  of  the  Guide  Pulley. 

It  is  a  well-known  fact  in  machine  design  that  a  belt  could  be 
run  over  the  pulleys  A  and  B,  shown  in  Fig.  73,  so  as  to  drive  in 
one  direction  or  the  other,  but  it  would  be  impossible  to  reverse 


116 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


the  direction  if  attached  to  run  in  one.  It  is  possible  to  construct 
an  "idler,"  or  guide  pulley,  with  its  shaft  at  such  an  angle  that 
the  belt  will  not  run  off  either  pulley,  running  in  either  direction. 


FIG.  74. — Locating  the  plane  of  the  guide  pulleys. 


FIG.  75. — Showing  revolution  and  counter-revolution  of  the  guide  pulley. 

Construction. — 1.  Select  a  point,  M  (Fig.  74),  between  the 
pulleys,  preferably  about  midway,  common  to  the  center  planes 
of  the  pulleys. 

2.  Run  a  tangent  from  M  to  each  of  the  pulleys  in  their  center 
plane. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     117 

3.  Draw  the  traces  of  X,  the  plane  of  the  two  tangents. 

4.  Revolve  the  tangents  intersecting  at  M  about  VX  (or  HX) 
and  construct  a  circle  of  any  suitable  dimensions  tangent  to 
them,  and  counter-revolve  it  to  show  its  H-  and  F-projections  in 
regular  position. 

112.  EXERCISES 

Use  any  combination  of  the  following  sizes  of  pulleys,  and  draw  the  projec- 
tions of  a  guide  pulley  at  a  convenient  point.  Make  all  pulleys  8-in.  face, 
and  place  the  working  pulleys  in  the  centers  of  their  respective  shafts.  Any 
reasonable  scale. 

Large  pulleys,  24  in.,  28  in.,  32  in.,  36  in.,  42  in.,  48  in.,  54  in.,  60  in 

Small  pulleys,  8  in.,  10  in.,  12  in.,  14  in.,  16  in.,  18  in. 

Guide  pulleys,  40  in.,  44  in.,  48  in.,  54  in.,  60  in. 

625.  Draw  the  projections  of  a in.  guide  pulley  to  run  between  a in. 

pulley  on  shaft  A(l^  -  K  +  2%)  B(1M  -  IK  +  2^)  and  a in. 

pulley  on  shaft  C(4  -  2  +  2)  D(4  -  2  +  1). 

626.  Same  for  shafts  E(l  +  2  -  %)  F(2  +  2  -  %)  and  G(4  +  1^-2) 
K(4  +  Y2  -  2). 

627.  Same  for  shafts  M(1K  +2-2)  N(1K  +1-2)  and  O(4  +  1^  - 

y2)  p(4  +  iy2  -  IK). 

628.  Same  for  shafts   A(l  -  1  +  1)    B(l  -  2  +  1)   and   C(2^  -1+2) 
D(4^  -  1  +  K). 

629.  Same    for    shafts    E(l  -  2,    0)    F(3  -  2  +  3)    and    G(4  -  Y2  +  1) 

K(4  -  iy2  +  i). 

630.  Same  for  shafts  L(l  +  3  -  K)  M(l  +  Y2  -  3)  and  N(4  +  1  -  K) 
O(4  +  3  -  3). 

631.  Same  for  shafts  P(l  +  3  -  Y2)  A(l  +  Y2  -  3)  and  X(4^  +  2  -  2M) 
Y(5M  +  2  -  2Y2}. 

632.  Same  for  shafts  B(M  +1-1)  D(l^  +1-1)  and  E(4  +  2  -  1) 
K(4  +  2  -  3). 

113.  THE  PROBLEM  OF  REFLECTION 

Problem  38. — To  find  the  points  at  which  a  ray  is  reflected 
from  planes  hi  passing  through  two  points  in  space. 

The  basis  of  this  solution  is  the  law  of  physics,  that  the  angle 
of  incidence  is  equal  to  the  angle  of  reflection. 

Refer  to  Fig.  76.  A  ray  is  to  pass  through  A,  strike  the  plane 
CC,  and  reflect  to  B.  To  find  D,  the  point  of  reflection. 

Imagine  CC  a  mirror.  The  observer  at  A  looks  at  the  mirror 
in  the  direction  of  B',  which  is  the  reflection  of  B,  and  is  located 
an  equal  distance  on  the  opposite  side  of  the  mirror;  i.e.,  B'E  = 
BE.  The  line  of  vision  is  a  straight  line  from  A  to  B',  piercing 
the  plane  (mirror)  at  D.  The  angles  BDC  and  B'DC  are  equal, 
and  equal  to  ADC,  hence  a  ray  from  A  is  reflected  at  D  to  B. 


118 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


If  the  number  of  planes  is  increased,  the  straight  line  still  persists 
if  the  object  is  transposed  symmetrically  about  each  of  the  planes. 


FIG.  76. 

Construction. — Let  it  be  required  to  find  the  points  on  H  and 
V  where  a  ray,  through  A,  striking  both,  is  reflected  to  B  (Fig. 
77). 

1.  Locate  the  projections  of  B  at  bivbiH,  symmetrically  placed 
on  opposite  sides  of  H  and  V. 

2.  Draw  a  straight  line  from  A  to  bivbiH. 

3.  Locate  the  H-  and  F-piercing  points  of  this  line,  C  and  D. 


FIG.  77. 


4.  As  C  is  the  first  point  reached  it  is  the  point  of  reflection 
on#. 

5.  Transpose  div  to  dv,  an  equal  distance  above  H,  and  it  will 
be  the  point  of  reflection  on  V. 

6.  Draw  the  line  ACDB,  and  it  will  be  the  path  of  the  ray  from 
AtoB, 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     119 


114. 


PRACTICAL  EXERCISES 


633.  Find  the  length  of  all  the  oblique  rafters  in  Fig.  78.     Scale  ^  in.  =1  ft. 

634.  Find  the  true  angles  between  the  various  oblique  rafters  and  the 
horizontal  ridges  that  intersect  them  (Fig.  78). 

635.  Measure  all  the  valley  angles  between  the  planes  A  and  C,  A  and  D,  etc., 
in  the  roof  (Fig.  78). 


FIG.  78. 

636.  A  rectangular  chimney  3  ft.  6  in.  X  5  ft.  4  in.  is  to  be  put  through  roof 
C  in  Fig.  78.  Draw  its  H-  and  Y-projections,  and  the  true  size  of  the 
opening  in  the  roof. 


b. 


Windows  2'0"x3'6'; 

FIG.  79. 


637.  Roof  in  Fig.  78.    Place  window 


(Fig.  79)  in  the  center  of  roof 


638.  Cut  an  opening  4  ft.  square  in  roof ,  Fig.  78. 

639.  An  octagonal  stone  tower,  18  ft.  in  diameter,  and  40  ft.  high,  is  sur- 
mounted by  a  pyramidal  cap,  12  ft.  high.     Draw  its  projections,  and 

draw  in  the  center  of  each  of  the  four  visible  faces  window (Fig. 

79).     Draw  only  half  of  the  plan  and  the  elevation  of  the  cap.     Scale 
Y±  in.  =  1  ft. 

640.  A  14-in.  steam  feed  pipe,  whose  center  line  is  A(l  ft.  +  3  ft.  3  in.  — 
3  ft.)  B(4  ft.,  0-1  ft.).     The  plane  of  the  roof  shed  is  T(5  ft.  +  2  ft.) 
6  in.  (2ft.  —  3  ft.).     Draw  the  projections  of  the  opening  in  the  roof, 
and  its  true  size.     Determine  the  dip  of  the  roof.     Scale  1  in.  =1  ft. 


120 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


641.  A  trapeze  bar,  3  ft.  long,  is  suspended  by  ropes  3  ft.  long.     Find  how 
high  it  is  raised  by  twisting  through °  (60°,  75°,  90°,  105°,  120°). 

642.  What  angles  does  the  pipe  in  Ex.  640  make  with  H,  V,  and  T.     If  it 
keeps  its  direction,  what  is  the  size  of  the  hole  in  the  floor? 

643.  A  conical  tower  is  15  ft.  in  diameter  and  14  ft.  high.     Draw  two  lines 

in  the  surface  from  the  apex  to  the  base,  making °  (30°,  45°,  60°) 

with  each  other. 

644.  Draw  a  cube  of  1%-in.  sides  on  H.     Draw  the  traces  of  a  plane  that 
shall  cut  a  regular  hexagon  from  the  cube.     Revolve  the  hexagon  to 
prove  that  it  is  regular,  and  calculate  the  angle  that  Q  makes  with  H . 

645.  A  building  28  ft.  6  in.  X  41  ft.  3-in.  ground  plan  is  covered  by  a  hip  roof 
having  a  central  ridge  18  ft.  long,  commencing  10  ft.  from  the  front 
of  the  house.     The  front  corner  rafters  are  inclined  45°  from  the 
horizontal.     Draw  the  plan  and  elevation  of  the  roof.     What  is  its 
height?     Measure  the  hip  angles.     What  are  the  angles  made  with  the 
horizontal  by  the  rear  corner  rafters? 


B.  C. 

Moulding      Sections. 
FlG.  80. 


646.  Two  pulleys  24  in.  X  8  in.  are  located  on  parallel  horizontal  shafts. 
The  center  line  of  one  is  A (4  ft.  +  6  ft.  6  in.  -  18  in.)  B(6  ft.  +  6  ft. 
6  in.  —  6  in.),  and  the  other  passes  through  C(6  ft.  +  2  ft.  —  5  ft.). 
The  first  pulley  is  midway  between  A  and  B.     The  belt  passes  through 
a  wall  whose  plane  is  S(10  ft.  +  4  ft.)  6  ft.  (2  ft.  -  4  ft.).     Draw  the 
projections  and  true  size  of  the  opening  in  the  wall  to  accommodate 
the  belt.     Allow  2-in.  clearance  at  all  points.     Scale  %  in.  =  1  ft. 

647.  Inscribe  a  sphere  in  the  tetrahedron  A (2,  0-1)  B(4,  0  —  1)  C(3^, 
0-3)  D(3^  +  2  -  13£). 

648.  Circumscribe  a  sphere  about  the  tetrahedron  in  Ex.  647. 

649.  Inscribe  a  sphere  in  the  tetrahedron  M(2  +  1M  -l)N(3+2— J£) 
0(3M  +  Ui  -  1M)  P(3  +  1  -  2). 

650.  Circumscribe  a  sphere  about  the  tetrahedron  in  Ex.  649. 

651.  Pass    a    plane    equidistant   from   A(2  +3-2)    B(3  +  1  —  1)   and 
C(5  +  3  -  Y2)  D(3>£  +1-3). 

652.  Draw  the  H-  and  F-projections  of  a  1^*5 -in.  hex  nut,  drilled  and  cham- 
fered, on  the  center  line  M(2  +  1  -  3)  N(4  +  2  -  1).     Full  size. 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     121 

653.  Determine  the  outline  of  a  moulding  cutter  to  cut  section ,  Fig.  80. 

The  cutter  is  mounted  on  an  arbor,  3-in.  diameter,  and  the  center  line 
of  the  arbor  is  43^  in.  from  the  table  of  the  machine.     Full  size. 

654.  Two  pipes,  C(2  +  1%,  0)  D(3^,  0-2)  and  E(2^,  0  -  Y2)  F(4  + 
2  —  1)  are  to  be  connected  by  the  shortest  possible  horizontal  pipe. 
Allow  2  in.  for  connections.     What  is  its  length  to  a  scale  of  1  in.  = 
1  ft.? 

655.  Draw  the  projections  of  a  2-in.  square  pyramid,  whose  hip  angles  are 
°  (105°,  120°,  135°,  150°). 

656.  At  a  point  of  outcrop  A  (.4  +  2  —  1)  the  strike  of  a  body  of  ore  is  south, 
50°  west.     The  dip  is  45°.     A  tunnel  is  driven  from  B(l  +  3  —  K), 
which  is  on  the  surface,  to  the  ore  perpendicularly.     Where  will  it 
strike  the  ore? 

Note. — The  "strike"  of  a  stratum  of  ore  is  a  horizontal  line  in  the  plane 
of  the  ore.     The  "dip"  is  the  inclination  of  the  ore  plane  to  the  horizontal. 

657.  The  point  of  outcrop  is  A(200  ft.  +  200  ft.  —  100  ft.),  and  a  line  is  run 
to  B,  a  distance  from  A  200  ft.,  south,  30°  west.     The  altitude  at  B  is 
125  ft.,  and  a  vertical  bore  hole  strikes  ore  at  a  depth  of  25  ft.     Another 
bore  hole  is  made  at  C,  on  the  same  level  as  B,  and  150  ft.  distant 
from  A  and  B.     At  C  the  hole  is  drilled  100  ft.,  striking  the  ore.     De- 
termine the  dip  and  strike  of  the  vein.     Scale  1  in.  =  100  ft. 

658.  Points  D,  E  and  F  are  on  the  side  hill,  which  dips  20°  east.     D  is  a 
point  of  outcrop  of  a  vein  of  ore.     From  D,  E  bears  south  45°  east 
400  ft.  on  the  slope  of  the  hill,  and  F  bears  south  15°  east  600  ft.  on 
the  slope  of  the  hill.     Vertical  drill  holes  at  E  and  F  strike  ore  at  100 
ft.  and   150  ft.  respectively.     Determine  the  dip  and  strike  of  the 
vein.     Scale  1  in.  =  100  ft. 

659.  At  A (8  ft.  +  24  ft.,  0)  and  B(20  ft.  +  24  ft.,  0)  in  a  mill  wall,  two 
parallel  6-in.  pipes  run  downward  to  the  right  at  45°  to  the  floor  and 
30°  to  the  wall.     It  is  proposed  to  run  a  12-in.  cylindrical  chute  on 
the  center  line  C(30  ft.  +  16  ft.,  0)  D(4  ft.,  0  -  24  ft.).     Can  this  be 
done?     If  not,  how  much  could  D  be  shifted  toward  the  wall  to  effect 
it?     Scale  H  in.  =  1  ft. 

660.  A  pulley,  24-in.'  diameter,  and  12-in.  face,  runs  on  the  shaft  A  (.2  ft. 
+  2  ft.  -  6  ft.)  B(10  ft.  +  7  ft.  -  2  ft.).     Draw  the  projections  of  the 
pulley.     Scale  ^  in.  =1  ft. 

661.  Draw  the  projections  of  a  IJ^-in.  sphere  tangent  to  the  plane  T(l  +  3) 
4(1  -  2)  at  the  point  O (3  +  34,  x)  in  T. 

662.  Measure  the  various  lineal  and  dihedral  angles  of  a  hexagonal  pyra- 
mid of  2-in.  base  and  2^-in.  altitude. 

663.  Draw  the  projections  of  a  10-ft.  circular  skylight  in  a  roof  whose  plane 
is  S(8  ft.  +  24  ft.)  32  ft.  (8  ft.  +  16  ft.),  with  its  center  at  M(20  ft.  + 
8  ft.,  x)  in  the  roof.     Scale  %  m-  =  1  ft. 

664.  The  eaves-line  of  a  roof  is  A(16  ft.,  0-20  ft.)  B(33  ft.,  0-5  ft.). 
The  horizontal  ridge  is  13  ft.  6  in.  high.     The  rafter  at  A  runs  perpen- 
dicularly to  AB,  dipping  60°.     The  rafter  at  B  runs  parallel  to  V  up  to 
the  ridge.     Calculate  the  cost  of  covering  this  roof  at  $1.50  per  square 
yard.     Scale  ^  in.  =1  ft. 


122 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


665.  A  tunnel  through  a  hill  runs  from  A (425  ft.  +  275  ft.  -  100  ft.)  to 
B(2100  ft.  +  150  ft.  -  1300  ft.).     Find  its  length  and  grade. 

666.  A  plane  T  passes  through  GL  and  the  point  K(3  +  1  —  2).    A  plane 
S  parallel  to  it  cuts  V  \Y±  in.  above  GL.     Find  the  H-trace  of  S,  the 
distance  between  T  and  S,  and  their  angles  with  H  and  V. 

667.  Pass  a  plane  through  L(l  +2  —  1)  equally  distant  from  M(4  +  1  — 
%),  N(3  +  2  -  H),  and  O(2  +  2  -  ^).     Find  a  point  in  the  plane 
thus  found  which  is  equally  distant  from  M,  N  and  O. 

668.  Erect  a  cube  on  H  with  its  base  diagonal  A(3,  0  -  K)  B(4,  0  -  1^)- 
Show  its  picture  as  seen  from  C(3K  +  1%  ~  3).     Show  the  picture  in 
its  true  size. 

Note. — Pass  the  picture  plane  perpendicular  to  the  line  from  the  point  of 
sight  to  the  center  of  the  cube. 

669.  From  a  hole,  A(18  ft.  +  10  ft.,  0),  in  a  mill  wall,  run  a  pipe  that  will 
make  a  straight  connection  with  two  pipes,  whose  center  lines  are 
B(5  ft.  +  10  ft.  -  3  ft.  6  in.)  C(19  ft.,  0-10  ft.)  and  D(3  ft.,  0-8 
ft.)  E(10  ft.  +  10  ft.  -  8  ft.).     What  are  the  lengths  of  the  two  por- 
tions of  the  pipe?     Scale  Y±  in.  =1  ft. 


FIG.  81. — Stone  arch. 


670.  Draw  the  projections  of  the  circle  passed  through  A (2  +2  —  J^), 
B(2K  +1-D,  and  C(3M  +  2  -  1^). 

671.  The  hinge  of  a  door  is  M(4  +  3,  0)  N  (4,  0,  0).     When  shut,  it  is  to  the 
left  of  MN.     A  taut  wire  runs  from  O(3  +  2,  0)  to  P(5,  0  -  1).    Find 
the  outline  of  a  slot  that  will  allow  the  door  to  open  freely  through  90°. 

672.  Let  H  be  a  mirror.     A  ray  of  light  passes  through  A(l  +  1^  —  2J^) 
strikes    H   and    passes    through    B(3%  +  1  —  M)-     Where    does    it 
reflect  on  H,  and  where  does  it  strike  V? 

673.  Find  the  path  of  a  ray  of  light  that  passes  through  O(5  +2  —  IK) 
and  is  reflected  from  H  and  V  through  P(2  +  1  —  2). 

674.  The  top  of  a  large  bin  in  a  mill  has  for  its  corners  the  points  A  (4  ft.  + 
7  ft.  6  in.,  0),  B(20  ft.  +  7  ft.  6  in.,  0),  C(4  ft.  +  2  ft.  3  in.  -  8  ft. 
9  in.),  and  D(20  ft.  +  2  ft.  3  in.  -  8  ft.  9  in.).     From  M(16  ft.  +  12 
ft.,  0)  the  center  line  of  a  24-in.  square  chute  runs  downward  to  the 
left  at  45°  to  H  and  30°  to  V.     Draw  the  projections  of  the  hole  in  the 
bin  top  made  by  the  chute,  and  find  its  true  size.     Scale  K  m-  =1  ft- 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     123 


675.  Draw  the  projections  of  four  spheres,  1^  in.,  1^  in.,  1%  in.,  and  2  in. 
in  diameter,  each  one  tangent  to  the  other  three. 

676.  Develop  the  surface  of  the  keystone  of  the  arch  shown  in  Pig.  81.     The 
masonry  is  14  in.  thick.     Use  any  convenient  scale. 

677.  M(8  in.  +  7>£  in.  -  2  in.)  N  (17  in.  +  4  in.  -  8  in.)  is  the  center  line 
of  a  hollow  tile,  4  in.  X  12  in.,  of  square  section  with  rounded  corners, 
M-in-  walls.     The  rounded  corners  are  1%-in.  radius.     Draw  its  pro- 
jections, scale  3  in.  =1  ft. 

677.  A  borehole  cuts  a  6-ft.  core  from  a  vein,  whose  strike  is  north  60° 
east,  and  whose  dip  is  45°  to  the  northwest.  The  borehole  bears 
south  75°  west  and  dips  30°.  What  is  the  real  thickness  of  the  vein? 


\B 

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Note:- 

•  Cross-hatched  por- 
tion of  P/art  is  H  pro-    1  1  ' 
jection  of  B  face        \  \ 

Ex.  681.— Plate  band  arch. 

678.  At  A,  where  the  elevation  is  9800  ft.,  a  vertical  borehole  strikes  a  vein 
at  300  ft.     B  bears  north,  30°  east,  1500  ft.  from  A,  and  has  an  ele- 
vation of  9600  ft.     An  inclined  hole  is  drilled,  bearing  north  75°  east, 
dipping  75°,  striking  the  vein  at  525  ft.     At  C,  whose  elevation  is 
9900  ft.,  1200  ft.  from  A,  south  60°  east,  a  vertical  borehole  cuts  the 
ore  at  375  ft.     Required  the  dip  and  strike  of  the  vein.     Let  H  be  at 
9500  ft.  level. 

679.  At  A(l  +  3  —  3)  a  borehole  is  drilled,  bearing  north  30°  east,  and 
dipping  75°.     It  intersects  a  vein,  whose  strike  is  due  north,  and  cuts 
an  8-ft.  core.     The  true  thickness  of  the  vein  is  5  ft.     What  is  the  dip 
of  the  vein? 

680.  Vertical  holes  on  a  hill  side  (considered  a  plane)  are  bored  150  ft.  at 


124 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


A(100  ft.  +  400  ft.  -  300  ft.),  200  ft.  at  B(350  ft.  +  250  ft.  -  325  ft.), 
and  125  ft.  at  C(600  ft.  +  325  ft.  -  100  ft.),  striking  a  vein  of  ore. 
What  is  the  line  of  outcrop,  and  what  are  the  dip  and  strike  of  the 
vein? 

681.  Draw  the  projections  of  the  Plate  Band  Arch.     Lay  out  the  patterns 
for  joints  A,  B,  C,  D,  E.     Scale  Y^  in.  =  1  ft. 

682.  Plate  Band  Arch.     Lay  out  the  patterns  for  the  slanting  faces,  F,  J, 
K,  L,  M,  and  N.     Scale  %  in.  =  1  ft. 

683.  Plate  Band  Arch.     Make  working  drawings  in  three  views  of  Block 

-  (A,  B,  C,  N,  M).     Scale  1  in.  =  1  ft 

Leo  Shaft 


Bench 

Mark 


Datum  Plane  Level  of  Shaft  Collar 


7S  Deep 


68'4, 


Outcrop  of  Lode 
Q  dips  [downward) 
60'onN.W.sic/e. 
Outcrop  of  Lode 
P  dips(downward) 
30° on  SE.  side 

Ex.  684. — Mine  survey  of  Prince  and  Queen  veins. 

Given  the  outcrops  of  Prince  and  Queen  veins,  and  the  shaft  Leo. 
Required  the  length,  bearing,  and  inclination  of  the  shortest  cross- 
cut from  the  present  bottom  point  of  the  shaft  to  the  intersection  of  P 
and  Q.  Also  the  length,  bearing,  and  level  of  the  drift  from  the  lowest 
point  of  this  line  to  the  line  of  the  shaft. 

Note. — Draw  compass  mark  and  scale. 
Mine  Survey 

9      .190          290 


Ex.  685. — Mine  survey  No.  2. 

685.  Mine  Survey  of  Anchoria,  Leland,  and  Isabella  lodes.     Required:  (1) 
Location  and  depth  of  a  shaft  to  the  intersection  of  A  —  L  —  I;  (2) 


PROBLEMS  RELATING  TO  POINTS,  LINES,  AND  PLANES     125 

length  and  bearing  of  a  10  per  cent,  tunnel  from  "location  x"  to  the 
intersection  of  A  and  L;  (3)  length,  bearing,  and  position  of  portal  on 
the  outcrop  of  Anchoria  to  the  intersection  A  —  L  —  I,  when  the  tun- 
nel has  a  10  per  cent,  grade. 

Note. — Print  Mine  Survey  like  plate  mark.     Draw  compass  mark  and 
scale.     Give  a  printed  statement  of  the  answers. 

Mine  Survey 

x  Dv  NM         N 

v  w-£-e 

soon        X S 

Level 

SMX  XN" 


X 
Ex.  686. — Mine  survey  No.  3. 

686.  Given  the  Shaft  Collar  of  a  winze  "S,"  and  a  hoisting  drum  at  "D." 
Required:  The  bearing  and  inclination  of  the  shaft  of  a  guide  pulley 
for  a  mine  cable  at  "N." 


CHAPTER  III 

INTERSECTIONS  AND  DEVELOPMENTS  OF  PLANE 

SOLIDS 

115.  Problem  39. — To  find  a  plane  section  of  a  pyramid  or 
prism. 

Analysis. — 1.  Obtain  the  piercing  points  of  the  various  oblique 
edges  with  the  given  plane. 


Fia.  82. 

2.  Connect  the  points  two  and  two  by  straight  lines  in  order. 
Construction. — Let  it  be  required  to  draw  the  section  line  of 
the  hexagonal  pyramid  made  by  the  plane  T  (Fig.  82). 
1.  Draw  the  Q-projection  of  the  pyramid. 

126 


INTERSECTION  AND  DEVELOPMENTS  OF  PLANE  SOLIDS     127 

2.  Draw  the  Q-trace  of  T. 

3.  Project  the  various  piercing  points  to  H  and  V.     (The 
Q-projections  of  these  points  are  all  located  on  QT.) 

4.  Join  the  points  in  order,  forming  the  irregular  hexagon 
enclosing  the  crosshatched  space. 

DEVELOPMENT 

• 

116.  Definition. — To  develop  the  surface  of  a  solid  is  to  bring 
its  various  faces  and  bases  into  one  plane,  all  in  their  true  sizes 
and  relative  positions,  so  that  the  plane  figure  thus  outlined  can 
be  cut  out  and  folded  into  the  form  of  the  solid. 

Note. — In  sheet  metal  drafting  this  is  called  "laying  out  the  pattern." 

Problem  40. — To  develop  the  surface  of  a  truncated  prism. 
Construction. — Let  it  be  required  to  develop  the  square  prism 
(Fig.  83),  between  its  fl-base  and  the  T-section. 


True  Size  of 
Oblique  Section 


FIG.  83. 


1.  Lay  out  the  "base  line"  on  a  right  section  of  the  prism. 
(In  this  case  use  the  fl-base.) 

2.  Lay  off  the  true  length  of  the  base  edges  on  this  line. 

3.  Erect  perpendiculars  at  these  points,  making  them  equal  in 
length  to  the  segments  of  the  longitudinal  edges,  in  order. 

4.  Connect  the  ends  of  these  perpendiculars  two  and  two  by 
straight  lines. 


128 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


5.  Attach  the  top  and  bottom  bases,  A  and  B,  to  their  proper 
edges  in  their  true  sizes. 

117.  Problem  41. — To  develop  a  truncated  pyramid. 

Construction. — Let  it  be  required  to  develop  the  pyramid 
shown  in  Fig.  82  between  the  If -base  and  the  T-section. 

1.  Obtain  the  true  lengths  of  the  edges,  and  of  the  segments 
between  bases. 

|y     2.  Lay  out  the  triangles  in  true  size  and  order,  with  a  common 
vertex. 


FIG.  84. — Development  of  truncated  pyramid. 

3.  Lay  off  on  the  edges  in  this  figure  the  true  lengths  of  the 
various  segments,  and  connect  these  points  by  straight  lines, 
two  and  two. 

4.  Attach  the  bases  A  and  B  in  their  true  sizes  to  the  proper 
edges. 


118. 


EXERCISES 


Graphic  Layout  No.  46. — Center  lines  for  regular  solids. 


INTERSECTIONS  AND  DEVELOPMENTS  OF  PLANE  SOLIDS     129 


f^ij^fira   Q  C)  Q  Q 

(0            (2)           13)           (4)             (5)          (6)           (7)          (8) 

t 

*\ 

3               / 

7 

4 

5 

V 

HW 

V£ 
HX 

K 

A 

// 

-w- 

7 

7 

8>*^ 

9| 

g 

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H 

11 

12     .^ 

vb" 

yz  
~z"~ 

-<- 

^ 

J 
1 

Graphic  Layout  No.  47.— Bases  for  regular  solids,  and  oblique  planes  for 
truncating.  In  laying  out  the  work  place  the  planes  so  that  they  will 
truncate  the  solids. 

Note. — If  desired  by  instructor,  the  layout  may  be  drawn  in  III. 


683.  Draw  the  intersection  of  the  regular  prism,  base ,  altitude  3  in., 

center  line ,  made  by  the  plane .      Find  the  true  size  of  the 

intersection. 

684.  Develop  the  surface  of  the  prism  in  Ex.  683  between  the  right  base  and 
the  oblique  base. 

685.  Pass  a  profile  plane  through  the  prism  specified  in  Ex.  683,  and  find 
the  P-projection  of  the  section. 

686.  Draw  the  intersection  of  the  regular  pyramid,  base ,  altitude 

in.  (2,  2%,  3),  center  line ,  made  by  the  plane .     Find  the  true 

size  of  the  section. 

687.  Develop  the  surface  of  the  pyramid  in  Ex.  686  between  the  right  base 
and  the  truncated  section. 

688.  Develop  the  surface  of  the  pyramid  in  Ex.  686  between  the  apex  and 
oblique  base. 

689.  Make  a  profile  section  of  the  pyramid  in  Ex.  686. 

690.  Make  a  full  set  of  details  of  all  the  glass  and  sheet  iron  pieces  in  the 
lamp  shade  and  supports,  shown  in  Fig.  85.     Do  not  duplicate  the 
details,  but  state  the  number  required  for  each.     Allow  for  clearance 
and  the  folding  of  the  iron  pieces. 

9 


130 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


FIG.  85. 


-- -  3'6 - 


Fia.  86. — Saw  horse. 


INTERSECTIONS  AND  DEVELOPMENTS  OF  PLANE  SOLIDS      131 


691.  Make  a  detail  of  the  backbone  of  the  saw  horse  shown  in  Fig.  86, 
showing  the  angles  of  the  cuts  to  be  made.  Draw  to  any  convenient 
scale. 


.         V 

FIG.  87. — Grain  hopper. 


Fia.  88.— Ventilator  cap. 


692.  Make  detail  drawings  of  the  legs  and  braces  of  the  saw  horse,  shown  in 
Fig.  86,  showing  the  angles  of  the  various  cuts.  Draw  to  any  con- 
venient scale. 


\ 

/ 

i 

1 
1 

XN*  -                 --/' 

\- 
\ 

'T 

TV--                      ^i 

\ 

IT"                            T 

ii                               II 
M                               i   I 

I  j_                           J   i 

\ 

\ 

\ 

s  •' 
«0 

y.   -                   ^ 

i    ^. 

/ 

\ 

\— 

i 
£ 

/' 

\ 

i 

|<  .....  .....  -  .......  - 


18"- 


14.' _.._ 


~ ->i\ 


T 


FIG.  89.— Stand. 


693.  Make  a  pattern  to  any  convenient  scale  of  the  grain  hopper,  shown  in 
Fig.  87.     Use  two  spaces  for  the  projections  and  pattern.     Make  a 
paper  model  of  the  hopper,  putting  on  a  flap  for  holding  together. 

694.  Make  a  pattern  to  any  convenient  scale,  of  the  ventilator  cap,  shown 
in  Fig.  88. 


132 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


695.  Fig.  89  shows  a  stool,  or  plant  stand,  made  of  1-in.  lumber.  The 
edges  of  the  four  inclined  supporting  pieces  are  chamfered  to  fit. 
Make  a  working  drawing  with  full  details  to  scale,  showing  the  angles 


FIG.  90. — Stone  arch. 

of  the  chamfered  pieces,  without  duplicating  any  of  the  parts.     Take 
two  spaces  for  the  work. 

696.  Develop  the  surface  of  the  keystone  of  the  arch  shown  in  Fig.  90.     The 
masonry  is  12  in.  thick.     Use  any  convenient  scale. 


CTC 


FIG.  92. 


119.  Problem  42. — To  find  the  line  of  intersection  between  two 
plane  solids. 

Construction. — To  find  the  line  of  intersection  between  the 
square  pyramid  and  square  prism,  shown  in  Fig.  91. 


INTERSECTIONS  AND  DEVELOPMENTS  OF  PLANE  SOLIDS     133 

The  line  of  intersection  between  two  plane  solids  is  a  broken 
line  (sometimes  two)  made  of  the  lines  of  intersection  between 
the  various  planes  bounding  the  solids. 

1.  Find  the  intersection  of  the  plane  ABEF  (the  upper  face 
of  the  prism)  with  the  pyramid.     This  is  the  quadrilateral  1234. 

Note. — It  is  not  necessary  to  find  the  traces  of  the  plane  to  accomplish 
this. 

2.  Find  the  intersection  of  the  plane  CDGK  (the  lower  face 
of  the  prism)  with  the  pyramid. 

3.  Connect  the  points  thus  found  with  the  limiting  points 
on  the  two  solids,  in  order,  and  the  resulting  broken  line  will  be 
the  line  of  intersection.     In  Fig.  91  this  is  the  heavy  line  in  each 
projection. 

120.  Second  Method. — By  projecting  on  an  auxiliary  plane. 
Construction. — To  find  the  line  of  intersection  between  the 
hexagonal  prism  and  the  triangular  prism,  Fig.  92. 


•  FIG.  93. 


Refer  to  Fig.  93.  Here  the  solids  are  projected  on  Q,  which  is 
taken  perpendicular  to  the  center  line  of  the  oblique  solid. 
Such  piercing  points,  as  cannot  be  obtained  from  the  H-  and  V- 
projections,  will  be  obtained  by  direct  projection  from  the  Q-pro- 
jection  to  the  //-projection  and  F-projection. 


134 
121. 


a 


18 


B  L|_i_i. 


y 

/ 


B 


Graphic  Layout  No.  48. — Center  lines  for  intersecting  solids.  Letter  to 
suit,  locating  in  I  or  III.  Another  pair  (9)  can  be  made  like  (8)  with  axes 
intersecting. 


3* 

>i  > 


Q 


Q 


H 

klcv 

Ji 


Graphic  Layout  No.  49.  —  Bases  for  solids. 


698.  Let  A 
tude 


,  alti- 


-  be  the  center  line  of  a  right  pyramid,   base 

in.  (2,  2^,  3),  and  B  —   —  be  the  center  line  of  the  right 

pyramid,  base ,  altitude in.  (2,  2^>  3). 

(o)  Draw  the  line  of  intersection. 

(6)   Develop  the  surface  of  pyramid  A. 

(c)    Develop  the  surface  of  pyramid  B. 
699.  Let  A  —   —  be  the   center  line  of  a  right  pyramid,   base 

tude in.  (2,  2%,  3),  and  B  —    —  be  the  center  line  of  the  right 

prism,  base ,  altitude  3  in. 


,  alti- 


INTERSECTIONS AND  DEVELOPMENTS  OF  PLANE  SOLIDS     135 

(a)  Draw  the  line  of  intersection. 

(6)   Develop  the  surface  of  pyramid  A. 

(c)   Develop  the  surface  of  prism  B. 

700.  Let  A be  the  center  line  of  a  right  prism,  base  ,  altitude 

2  in.  and  B  —   —  be  the  center  line  of  a  right  pyramid,  base  , 


altitude 


in.  (2,  2K,  3). 


701, 


(a)  Draw  the  line  of  intersection. 
(6)   Develop  the  surface  of  prism  A. 
(c)   Develop  the  surface  of  pyramid  B. 

Let  A be  the  center  line  of  a  right  prism,  base ,  altitude 

2  in.,  and  B be   the   center  line   of  a  right   prism,  base , 

altitude  3  in. 


(a)  Draw  the  line  of  intersection. 
(&)  Develop  the  surface  of  prism  A. 
(c)   Develop  the  surface  of  prism  B. 
702.  Make  a  paper  model  of  the  two  given  solids. 

CONCRETE  FORM  PROBLEMS 


Form  A 


Form  B 


FIG.  95. 


A                   B 

C 

D               E 

F 

s-\ 

H 

K 

ft.                  ft. 

ft. 

ft.  in. 

ft.  in. 

ft.  in. 

ft.  in. 

ft.  in. 

ft. 

20                   8 

16 

3 

4 

4     6 

2     6 

3 

8 

18                   7 

15 

2     9 

3 

5 

3 

2     9 

6 

16                   6 

13 

2     6 

2     6 

3 

2     9 

2     6 

.    5 

12                    5 

9 

2 

2                3 

2 

2     3 

4 

703.  Develop  the  patterns  for  the  concrete  form 


(A  or  B),  dimensions 


(1,  2,  3,  or  4),  constructed  by  the  Santa  Fe  Railroad  (dimen- 


sions altered). 


136 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


704.  In  laying  out  the  timbering  for  the  gable,  shown  in  Fig.  94,  use  sizes 
selected  from  the  following  list: 


4  in. 

X 

6  in. 

6  in. 

X 

6  in. 

Sin. 

X 

Sin. 

10  in. 

X 

10  in. 

4  in. 

X 

Sin. 

6  in. 

X 

Sin. 

Sin. 

X 

10  in. 

10  in. 

X 

12  in. 

4  in. 

X 

10  in. 

6  in. 

X 

10  in. 

Sin. 

X 

12  in. 

10  in. 

X 

14  in. 

4  in. 

X 

12  in. 

6  in. 

X 

12  in. 

Sin. 

X 

14  in. 

10  in. 

X 

16  in. 

Dimensions  of  the  gable  may  be  altered,  if  desired.  Detail  each  piece, 
laying  off  the  angles  of  the  trim,  with  full  details  of  measurements. 
Scale  to  suit. 


FIG.  94. — Timbering  of  a  gable  in  Main  Building,   Armour  Institute    of 

Technology. 

705.  Two  timbers,  whose  center  lines  are  A(4  ft.  +  10  ft.  -  2  ft.)  B(18  ft., 
0  -  5  ft.  6  in.)  and  C(4  ft.,  0-8  ft.)  D(18  ft.  +  10  ft.  -  8  ft.),  are 
connected  by  the  shortest  possible  timber  brace,  notched  to  fit.    Select 
the   timber   sizes  from  the  foregoing  list,  and  draw  the  projections 
of  the  three  pieces  in  assembly  and  detail,  with  full  dimensions  of 
notches  and  trims.     AB  and  CD  are  trimmed  to  fit  horizontal  timbers. 
Scale  Y±  in.  =  1  ft. 

706.  Using  the  same  data  as  Ex.  705,  make  the  crossbrace  the  shortest 
possible  horizontal  timber. 


INTERSECTIONS  AND  DEVELOPMENTS  OF  PLANE  SOLIDS     137 
RECTANGULAR  CONNECTION  PIPES 

i! 


A 

B 

C 

D 

E 

F 

G 

K 

in. 

in. 

in. 

in. 

in. 

in. 

in. 

in. 

6 

18 

6 

4 

10 

10 

4 

6 

5 

14 

5 

3 

X 

6 

3 

3 

4 

12 

4 

2 

6 

6 

3 

4 

4 

10 

3 

1 

6 

10 

2 

5 

A 

B 

C 

D 

E 

F 

G 

in. 

in. 

in. 

in. 

in. 

in. 

in. 

8 

3H 

4 

7 

8 

12 

8 

6 

3 

3 

7 

7 

10 

7 

5 

a 

2 

R 

6 

9 

6 

4 

2 

2 

5 

5 

8 

6 

TAPERED  CONNECTING  PIPE.  TWISTED  CONNECTING  PIPE. 

FIG.  96. 


707.  Develop  the  patterns  for  the  tapered  connection  pipe 


(Fig.  96). 


708.  Develop  the  patterns  for  the  twisted  connection  pipe (Fig.  96). 

709.  Develop  the  patterns  for  the  hopper (Fig.  97).     Any  convenient 

scale. 


A 

B 

C 

D 

e 

ft.in. 

ft.  in. 

ft. 

in. 

degrees 

4    2 

2   10 

o 

12 

15,  30,  45,  60 

3    8 

2     4 

5 

12 

15,  30,  45,  60 

3    2 

2     0 

4 

14 

15,  30,  45,  60 

2  10 

2     1 

4 

14 

15,  30,  45,  60 

2    6 

1     9 

4 

16 

15,  30,  45,  60 

HOPPER  FOR  SQUARE  CHUTE. 
FIG.  97. 


A 

B 

C 

D 

E 

e 

in. 

in. 

in. 

in. 

in. 

degrees 

10 

8 

2*4 

2 

12 

60,  75,  90 

10 

6 

2H 

IK 

12 

60,  75,  90 

10 

4 

2H 

1 

12 

60,  75,  90 

16 

12 

4 

3 

20 

60,  75,  90 

16 

10 

4 

2H 

20 

60,  75,  90 

16 

8 

4 

2 

20 

60,  75,  90 

24 

18 

8 

6 

30 

60,  75,  90 

24 

15 

8 

5 

30 

60,  75,  90 

24 

12 

8 

4 

30 

60,  75,  90 

ELBOW  HOPPER  (PYRAMIDAL). 
FIG.  98. 


138 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


710.  Develop  the  patterns  for  the  hopper  elbow  ,  in   (2,  3,  4) 

pieces  (Fig.  98).     Any  convenient  scale. 

711.  A  rectangular  air  pipe,  10  in.  X  12  in.,  follows  the  center  line  A (3  ft.  + 
6  ft.,  0)  B(3  ft.  +  6  ft.  -  2  ft.)  0(7  ft.  6  in.  +  2  ft.  -  2  ft.)  D(7  ft. 
6  in.,  0  —  2  ft.),  coming  through  the  wall  horizontally,  running  down- 


FIG.  99. 


ward  obliquely,  and  then  down  through  the  floor.     Keep  the  cross- 
section  uniform  except  at  the  turns.    Lay  out  the  joints  in  projection 
and  lay  out  the  patterns  for  the  piping.     Scale  ^  in.  =  1  ft. 
712.  Fig.  99  shows  a  45°  gable,  with  "Water  Table"  A  inclined  at  30°,  using 
moulding  "D,"  Fig.  100,  multiplying  each  dimension  by  3.     Required 


Moulding  Sections 
FIG.  100. 

the  line  of  intersection  of  the  moulding  with  the  water  table  at  points 
B  and  C.  Also,  lay  out  the  pattern  for  this  moulding  in  sheet -metal 
work.  (Note. — Develop  as  a  cylinder,  and  show  the  shape  of  each 
end  in  development). 


INTERSECTIONS  AND  DEVELOPMENTS  OF  PLANE  SOLIDS     139 

713.  Fig.  101  shows  a  45°  gable,  using  moulding  "B,"  Fig.  100,  multiplying 
each  dimension  by  4,  with  half-round  pilasters  as  shown  at  C  and  D. 
Draw  the  line  of  intersection  in  regions  E  and  F.  Lay  out  the  patterns 
for  this  moulding  in  sheet  metal.  Show  each  end  in  the  development. 
Lay  out  also  the  patterns  for  the  pilaster  in  sheet  metal.  Any  con- 
venient scale. 


FIG.  101. 


CHAPTER  IV 
SHADES  AND  SHADOWS 

122.  One  of  the  interesting  applications  of  Descriptive  Geome- 
try is  the  determination  of  the  shades  and  shadows  of  objects. 
Catalogue  cuts  and  architect's  drawings  employ  it  most  exten- 
sively, but  on  many  occasions  it  will  be  found  useful  in  shop 
drawings. 

123.  Definitions. 

Shade  is  that  part  of  an  object  not  exposed  to  the  direct  rays 
of  light. 


FIG.  102. 

Shadow  is  that  part  of  a  surface  in  light  from  which  rays  of 
light  are  excluded. 

Umbra  is  the  unlighted  space  behind  an  object  in  light. 

Note. — Shadow  is  the  intersection  of  the  umbra  with  a  surface  in  light; 
that  is,  it  is  the  interrupted  portion  of  an  umbra  by  an  interposed  surface. 

The  Line  of  Shade  is  the  imaginary  line  between  the  light  and 
shaded  portions  of  an  object. 

124.  The  sun  is  our  source  of  light,  and  its  rays  are  assumed 
to  be  straight  lines,  all  parallel.  This  assumption  is  very  nearly 
true,  and  the  deviation  of  the  rays  is  so  very  slight  that  no  ac- 
count need  be  taken  of  the  error.  For  convenience  in  conventional 

140 


SHADES  AND  SHADOWS  141 

shading,  the  sun's  rays  are  assumed  to  be  passing  over  the 
left  shoulder  of  the  observer  in  the  direction  of  the  diagonal  of 
the  conventional  cube;  that  is,  both  the  H-  and  F-projections  of 
the  rays  are  drawn  45°  to  GL.  What  angle  would  the  rays  then 
make  with  H  and  V? 

Referring  to  Fig.  102,  we  have  the  rays  of  light  depicted  as 
coming  toward  a  sphere  in  parallel  lines,  and  either  stopping  (be- 
ing reflected)  if  they  strike  the  sphere,  or  passing  on  to  the  next 
surface  in  light,  if  they  do  not.  One  half  the  sphere  is  in  light, 
the  other  half  in  shade,  and  the  line  of  shade  is  a  great  circle  of 
the  sphere.  This  shows  the  umbra  to  be  a  cylinder  of  equal 
diameter  to  the  sphere.  The  umbra,  on  meeting  another  surface, 
is  interrupted,  and  its  intersection  therewith  is  the  shadow. 

125.  Shadows  are  made  up  of  the  piercing  points  of  the  rays. 
From  the  foregoing  it  will  appear  that  the  composite  of  the 

tangent  rays  to  the  sphere  will 

pierce  the  surface  in  light  in  the  ° 

outline  of  the  shadow.     Hence,  |    \ 

defining  the  shadow  of  an  object  |        ^v 

resolves  itself  into  obtaining  the  \ 

tangent    rays   and    determining  \ 

their    piercing   points   with  the  ^pw 

various  objects  in  their  path.  | 

[  I 

/F~ 

126.  LAWS  / 

The  Shadow  of  a  Point  is  a  / 

Point.  / 

The  Umbra  of  a  Point  is  a  Line.  y£ 

The   Shadow  of  a  Line  is,  in     _       *      __      ,    , 

FIG.  103. — The  shadow  of  a  point, 
general,  a  Line. 

The  Umbra  of  a  Line  is,  in  general,  a  Plane. 
The  Shadow  of  a  Plane  Figure  cast  on  a  plane  is,  in  general,  a 
Plane  Figure  of  similar  outline. 

The  Umbra  of  a  Plane  Figure  is,  in  general,  a  Prism. 


THE  METHOD  OF  FINDING  SHADOWS 

127.  Problem  43.— To  find  the  shadow  of  a  point. 

Construction. — Let  O,  Fig.  103,  be  the  given  point. 

1.  Draw  the  ray,  so  that  ovpv  and  oHpH  each  make  45°  with  GL. 


142 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  Find  the  piercing  point  of  the  ray  with  H  or  7,  the  one  it 
meets  first. 

Conclusion.— O  casts  its  shadow  on  V  at  P. 


FIG.  104.— The  shadow  of  a  line. 


FIG.  105. — Shadow  of  a  line  on  H  and  V. 


128.  Problem  44.— To  find  the  shadow  of  a  line. 
Construction.— Let  AB,  Fig.  104,  be  the  given  line. 
1.  Find  the  piercing  points,  C  and  D,  of  A  and  B. 


SHADES  AND  SHADOWS 


143 


2.  Draw  CD;  it  is  the  shadow  on  H. 

Special  Case. — When  a  line  casts  its  shadow  on  more  than 
one  plane. 

Construction. — Let  AB,  Fig.  105,  be  the  given  line. 

1.  Find  its  /^-piercing  points,  C  and  D,  and  the  V-piercing 
point,  E,  of  B. 

2.  Draw  cHdH  until  it  meets  GL,  and  connect  it  at  this  point 
with  ev. 

Note. — The  shadow  line  is  the  trace  of  the  umbra  plane,  hence  this  shadow 
is  the  H-trace  on  H,  and  the  F-trace  on  V  of  the  plane  ABCD. 

129.  Problem  45. — To  find  the  shadow  of  a  plane  figure  on  a 
plane. 


avbv 


FIG.  106. — Shadow  of  a  square. 

Construction. — Let  the  square  ABCD  (Fig.  106)  be  the  figure; 
to  find  its  shadow. 

1.  Obtain  the  ff-piercing  points  of  the  rays  from  the  four 
corners,  A,  B,  C,  and  D,  at  eH,  fH,  gH,  and  hH. 

2.  Join  them  in  order,  and  fill  in  the  enclosed  space,  except 
where  it  is  covered  (hidden)  by  the  opaque  object. 

Note. — If  the  plane  figure  is  parallel  to  the  surface  in  light,  its  shadow 
will  be  an  equal  plane  figure.     Why? 

130.  Problem  46. — To  draw  the  shadow  of  a  solid  on  a  plane. 
Construction. — Let  the  cube  shown  in  Fig.  107  be  the  given 
solid. 

1.  Draw  rays  from  all  the  corners. 

2.  Disregard  the  ray  through  A,  as  it  will  be  inside  the  umbra. 


144 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


3.  Obtain  the  //-piercing  points  of  the  rays  from  B,  C,  and  D, 
at  eH,  fH,  and  gH. 

4.  Connect  these  points  with  the  broken  line,  and  fill  in  back 
to  the  H -projection  of  the  cube. 

Note. — The  shadows  of  BM  and  DN  are  the  lines  mHeH  and  nHgH. 


V.  V                                                                     "  IT 

a  b                                   cd 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

v                                          \ 

v                                           \    rv 

m                              \ 

e                                 \f 

FIG.  107. — Shadow  of  a  cube. 


Special  Case. — When  the  shadow  falls  on  more  than  one 
plane. 

Construction. — Let  the  hexagonal  prism,  Fig.  108,  be  the 
object.  To  find  its  shadow  on  H  and  V. 

1.  From  all  the  exposed  points  draw  the  rays  and  obtain  their 
piercing  points.     In  this  case  it  will  be  found  that  the  point  X 
is  the  only  one  to  cast  its  shadow  on  H . 

2.  Obtain  the  shadow  of  MX  on  H  and  V,  yH  -  nv,  and  of  AB 
on  H  and  V,  aH  —  CH  —  cv  (Problem  44,  Special  Case). 


SHADES  AND  SHADOWS 


145 


3.  Join  the  remaining  points  in  order,  thus  having  part  of  the 
shadow  on  H  and  part  on  V. 


131. 


FIG.  108. — Shadow  of  a  prism  on  H  and  V. 


GENERAL  REMARKS 


The  foregoing  examples  show  the  principles  and  rudiments  of 
the  subject  of  Shades  and  Shadows.  Curved  surfaces,  oblique 
planes;  and  complicated  objects  do  not  alter  the  principles,  and 
by  applying  the  principles  already  enunciated,  the  most  com- 
plex problems  can  be  worked  with  no  further  instruction. 


132. 


EXERCISES 


Draw  the  shades  and  shadows  of  the  following  objects,  on  the  speci- 
fied surfaces  in  both  projections  (and  profiles,  if  necessary).  Double- 
crosshatch  the  shadows,  single-crosshatch  the  shades,  and  use  dotted  cross- 
hatching  for  the  invisible  parts  of  the  shadows.  For  example,  see  Fig.  108. 

Note.  —  When  no  planes  are  specified  for  the  shadows,  H  and  V  are 
implied. 

714.  A  cube  on  H,  having  A(3,  0  —  1)  B(3,  0  —  3)  for  its  base  diagonal. 

715.  Twist  the  cube  in  Ex.  714,  -    -  °  (10°,  20°,  30°,  40°). 

716.  Oblique  cone,  1^-in.  circular  base  in  H,  axis  A  (2,  0  —  2)  B(4  +  2  —  1). 

717.  Hexagonal  pyramid,  2-in.  base,  center  line  C(2,  0  -  1M)  D(2  +  3  - 


718.  Pyramid  in  Ex.  717,  center  line  E(l  +1^-2)  F(3^  +  \Y±  -  2). 
10 


146 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


719.  Hexagonal  prism  resting  on  H,  center  line  G(l)^  +  %  —  1) 
X  -2). 

720.  Raise  the  prism  in  Ex.  719  K  in.  off  H. 

721.  Cylinder,  2-in.  diameter,  axis  M(2,  0  -  1^)  N(2  +  3  - 

722.  Surmount  the  cylinder  in  Ex.  721  with  a (square  or  hexagonal) 

plinth,  3-in.  diagonal  and  %  in.  thick. 

723.  Cylinder  tangent  to  H,  axis  O(l  +  %  -  1)  P(3  +  H  -  2). 

724.  Prism (square  or  hexagonal),  base  diagonal  2  in.,  center   line 

A(2,  0-2)  B(2  +  2  -  2).     Shadow  on  H  and  T(5  +  2)  2(5  -  2).  . 

725.  Prism in  Ex.  724  cutting  through  the  plane  S(  +  2)   «  (  -  3). 

Shadow  on  S. 

726.  Square  prism,  1-in.  base,  center  line  A(2,  0  —  2)  B(2  +  2  —  2),  sur- 
mounted by  a  square  plinth,  1%-in.  base,  J£  in.  thick. 


T" 

5 
-N 


V     _ 


FIG.  109. 


727.  Prism  and  plinth  in  Ex.  726.     Shadow  on  H,  V,  and  T(5  +  2)  2(5  -  2). 

728.  Prism   and  plinth  in   Ex.  726,  emerging  through  S(  +  2)    °°(  —  3). 
Shadow  on  S. 

729.  Hexagonal  prism,   1^-in.  base  diagonal,  center  line  A(2,  0  —  2)  B- 
(2  +  2  —  2),  surmounted  by  a  hexagonal  plinth,  2-in.  base  diagonal, 
%  in-  thick. 

730.  Prism  and  plinth,  Ex.  729.     Shadow  on  H,  7,  and  T(5  +  2)  2(5  -  2). 

731.  Prism  and  plinth,  Ex.  729,  cutting  through  plane  S(  +  2)   oo(  -  3). 
Shadow  on  S. 

732.  Cylinder    1-in.    diameter,    surmounted   by (square,    round,    or 

hexagonal)  plinth,  base  diagonal  (or  diameter)  2  in.,  M  m-  thick. 

733.  Cylinder  and  plinth,  Ex.  732.     Shadow  on   H,  V,  and  T(5  +  2) 
2(5  -  2). 


SHADES  AND  SHADOWS 


147 


734.  Cylinder  and   plinth,   Ex.  732,   cutting   through    S(  +  2)    »(  -3). 
Shadow  on  S. 


— K 


FIG.  110. 

<&. 

735.  Cross,  Fig.  109,  base  center  at  A(2,  0  -  1). 

736.  Twist  the  cross,  Ex.  735, °  (10°,  20°,  30°,  45°,  60°,  75°). 


FIG.  111. 

737.  Cross,  Ex.  736,  center  at  B(2,  0  -  2).     Shadow  on  H,  V,  and  T(5  +  2) 

2(5  -  2). 


148 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


738.  Place  the  cross,  Ex.  736,  on  a  stepped  pyramid  made  up  of  four  J£  in. 
thick  plinths,  respectively  1%  in.,  1^  in.,  1J4  in->  1  in.  square. 

739.  Draw  the  box,  Fig.  110,  30°  from  V.     Shades  and  shadows,  inside  and 
outside,  and  on  H  and  V. 

740.  Draw  the  box,  Fig.  110,  on  S(l  +  1)  3(3  -  3).     Shadows  on  H,  V 
and  S. 

741.  Draw  the  box,  Fig.  110,  on  end,  open  front.     Shadows  on  H,  V,  and 
itself. 

742.  Twist  the  box,  Fig.  741, °  (10°,  20°,  30°).     Shadows  on  H,  V,  and 

itself. 

743.  Side  elevation  of  a  house  with  45°  gable  roof.     Draw  plan  and  end 
view.     Required  shadow  of  chimney  on  roof,  roof  edges  on  walls, 
and  entire   house  on  the  ground.     Dimensions  and  scale  whatever 
convenient. 


FIG.  112. 

744.  Side  elevation  of  a  house  with  30°  gable  roof.  Draw  the  plan  and  end 
view.  Draw  the  shadow  of  the  chimney  and  dormer  on  the  roof,  also 
the  roof  edges  on  the  walls,  and  the  shadow  of  the  house  on  the  ground. 


CHAPTER  V 

CURVED  LINES 

133.  Definitions. 

A  Line  is  the  path  of  a  point. 

A  Straight  Line  is  a  line  in  which  the  point  moves  continually 
in  one  direction. 

A  Curved  Line  is  a  line  in  which  the  point  moves  in  a  direction 
that  is  being  changed  constantly  in  obedience  to  some  law. 

A  line  is  regarded  as  having  one  dimension  only — length, 
without  breadth  or  thickness.  It  is  considered  as  being  made 
up  of  infinitely  small  points,  separated  by  infinitely  small  spaces. 
The  line  may,  therefore,  be  regarded  as  being  generated  by  an 
immaterial  point,  occupying  consecutive  positions.  Two  of 
these  consecutive  points  determine  a  straight  line,  of  infinitesimal 
length,  but  of  a  determinate  direction.  Such  a  line  is  called  an 
Element;  that  is,  it  is  one  of  the  innumerable  infinitely  small 
straight  lines  that  compose  every  line,  straight  or  curved. 

134.  Classification  of  Lines. 
Lines  are  Straight  or  Curved. 

Curved  Lines  may  be  Plane  Curves,  also  called  Lines  of  Single 
Curvature,  or  they  may  be  Space  Curves,  also  called  Lines  of 
Double  Curvature. 

A  Plane  Curve  is  a  line  moving  always  in  a  plane,  changing 
its  direction  in  accordance  with  some  law.  Examples,  circles, 
ellipses,  etc. 

A  Space  Curve  is  a  curved  line,  no  four  consecutive  points  of 
which  are  contained  in  any  plane.  Example,  the  helix. 

Table  of  Lines 
Straight.     Only  one  kind. 
'  Circle 
Ellipse 


Plane  Curves 


.    ,       }   Conic  Sections. 
Hyperbola 

Parabola 


Involutes 

o     ,  • ,  ,  r,  (  Gearing  Curves. 

Cycloidal  Curves    ) 

Spirals,  etc. 
[  Helix 
Space  Curves     I  Most  Intersections  of  Curved  Surfaces 

[  Spherical  Epicycloid,  Spherical  Hypocycloid. 
149 


150 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


135.  Projections  of  Curves. 

A  curve  is  projected  on  H,  V  and  P  in  the  same  manner  as  a 
straight  line;  that  is,  all  the  points  in  the  curve  are  projected  on 
the  plane  of  projection  in  lines  perpendicular  to  the  plane,  and 
the  sum  of  these  projections  of  points  makes  up  the  projection  of 
the  line.  As  these  projectors  are  all  parallel,  the  surface  which 
they  compose  is  called  the  Projecting  Cylinder  of  the  curve. 
(Compare  with  the  projecting  plane  of  a  straight  line.) 


FIG.  113. — Projections  of  a  curve. 

Usually  a  small  number  (six  or  eight)  points  in  space  will  very 
accurately  determine  a  curve.  Its  projections  are  drawn  by 
running  a  smooth  line  through  the  projections  of  the  points  in 
order,  on  both  H  and  V,  and  on  P,  if  necessary.  (See  Fig.  113.) 

Note. — Plane  curves,  in  certain  positions,  will  project  on  one  plane  as  a 
straight  line,  and  the  projecting  cylinder  is  in  that  case  a  projecting  plane. 
However,  most  curves  do  not  lie  so  conveniently,  and  space  curves  never 
project  as  a  straight  line.  Any  curve,  that  is  parallel  or  oblique  to  the 
plane  of  projection,  will  project  as  a  curve  on  that  plane.  Any  plane  curve, 
lying  in  a  plane  perpendicular  to  the  plane  of  projection,  will  project  as  a 
straight  line  on  that  plane. 

136.  Plane  Curves  Lying  in  a  Profile  Plane. 

No  curve,  lying  in  a  profile  plane,  can  be  shown  in  its  true 
character  by  the  H-  and  F-projections,  because  they  will  both  be 


CURVED  LINES 


151 


straight  lines.  The  P-projection  will  show  not  only  the  true 
character  of  the  curve,  but  its  true  dimensions.  The  H-  and 
F-projections  of  a  number  of  points  in  a  curve  will  determine  it, 
but  will  not  show  it.  (See  Fig.  114.) 

In  a  similar  way,  any  curve,  lying  in  a  plane  parallel  to  H,  is 
projected  in  its  true  size  on  H,  and  as  a  straight  line  on  V. 

If  the  curve  lie  in  a  plane  parallel  to  V,  its  F-projection  gives 
its  true  size,  and  its  //-projection  will  be  a  straight  line. 

If  a  curve  lie  in  a  plane  perpendicular  to  H  and  oblique  to  V, 
its  //-projection  will  be  a  straight  line,  and  the  F-projection  will 
be  curved,  and  will  show  the 
character  of  the  curve.     The 
curve,    however,   may  be 
entirely  changed  by  this  fore- 
shortening; e.g.,  the  projec- 
tion of  an  oblique  circle  is  an 
ellipse. 

The  H-  and  F-piercing 
points  of  a  curve  are  found 
by  the  same  rule  that  obtains 
for  straight  lines;  that  is, 
where  the  F-projection 
crosses  GL,  project  to  the 
//-projection  to  find  the  H- 
piercing  point,  etc. 

137.  Tangents  and 
Normals  to  Curves. 

Definition. — A  tangent  to 
a  curve  is  a  line  containing 
two  consecutive  points  of  the       FlG-  114.— Curve  in  a  profile  plane, 
curve. 

The  tangent  may  be  straight  or  curved,  but,  when  curved,  it 
is  always  especially  designated.  When  spoken  of  as  a  tangent, 
the  understanding  is  that  it  is  a  straight  line  tangent. 

The  tangent  (from  its  definition)  must  lie  in  the  plane  of  the 
curve  at  the  point  of  tangency.  Also  (from  its  definition)  it 
must  have  the  same  direction  as  the  curve  at  the  point  of  tan- 
gency. It  is  especially  important  to  remember  that  the  tangent 
must  lie  in  the  plane  of  the  curve,  as  many  errors  occur  from 
disregarding  this  fact. 


152  PRACTICAL  DESCRIPTIVE  GEOMETRY 

Note. — The  apparent  discrepancy  between  the  foregoing  definition  of  the 
tangent  and  that  given  in  most  Plane  Geometries  amounts  to  nothing. 
They  define  a  tangent  as  a  line  touching  a  curve  at  one  point.  Their  defini- 
tion answers  sufficiently  for  beginners  in  Geometry,  and  for  a  section  of  the 
subject  that  confines  its  operations  to  a  single  plane,  but  the  student  can 
easily  see  how  such  a  definition  would  be  utterly  inadequate  in  Space 
Geometry. 

As  the  tangent  is  a  short  straight  line  in  the  curve  (two  con- 
secutive points),  the  curve  may  be  considered  as  being  composed 
of  an  infinite  number  of  infinitesimal  tangents,  intersecting  con- 
secutively. Fig.  115  shows  a  series  of  straight  lines,  A,  B,  C, 
etc.,  intersecting  at  short  finite  intervals.  They  compose  so 

nearly  a  curve  that  it  is  easy  to 
imagine  that  minute  divisions 
would  compose  an  actual  curve. 
From  the  figure  it  can  easily  be 
seen  how  the  tangent  at  any 
point  determines  the  direction 
of  the  curve  at  that  point. 

138.  Projections  of  the  Tan- 
gent. 

The  projections  of  the  tangent 
to  a  curve  will  be  tangent  to  the 

projections  of  the  curve  on  the  respective  planes.     The  converse 
is  not  necessarily  true,  because  a  line  lying  in  a  different  plane 
from  that  of  the  curve  may  have  its  projection  tangent  to  that 
of  the  curve  on  one  plane,  but  not  on  any  other. 
To  Draw  a  Tangent  to  a  Curve. 

There  are  exact  and  approximate  methods.     In  the  case  of  the 
circle,  the  tangent  is  always  perpendicular  to  the  radius  at  the 
point  of  contact.     In  the  case  of  other 
conic  sections,  the  exact  method  will  be  ^^   , 

taken  up  in  connection  with  the  study     /tJ 
of  those  curves.     For  ordinary  drafting 
work  a  sufficiently  accurate  tangent  may  FIG.  116. 

be    drawn  by  placing  a  straight   edge 

against  the  curve,  and  laying  off  the  tangent  with  the  eye.  A 
simple  approximation,  that  can  be  used  in  nearly  all  cases  of 
curves  and  is  exact  for  the  circle,  is  this:  Let  it  be  required  to 
draw  the  tangent  at  A  (Fig.  116)  on  the  curve  shown.  Set  the 
dividers  at  some  small  arbitrary  span,  and  lay  off  B  and  C  on 


CURVED  LINES  153 

the  curve  at  equal  distances  from  A.  Connect  B  and  C  with  a 
straight  line.  Through  A  draw  AD  (the  tangent)  parallel  to  BC. 
This  method  will  not  do  for  curves  changing  their  radii  rapidly. 
The  eye  will  do  close  work  in  sharply  curving  lines.  Numerous 
close  approximations  may  be  found  in  other  text-books,  and  an 
especially  good  one  is  given  in  MacCord's  "Elements  of  De- 
scriptive Geometry,"  page  93,  but  its  practical  value  is  small. 

139.  Normals. — A  normal  to  a  curve  at  any  point  is  a  perpen- 
dicular to  the  tangent  at  the  point  of  tangency.     While  any 
perpendicular  is  normal  to  the  curve,  it  is  customary  to  regard 
the  normal  as  the  perpendicular  to  the  tangent  in  the  plane  of 
the  curve. 

In  the  case  of  a  space  curve,  the  normal  may  be  any  line  per- 
pendicular to  the   tangent   at 
the  point  of  tangency. 

140.  The    Rectification    of 
Curves. 

To  rectify  a  curve  means  to 
straighten  it  out ;  that  is,  to  draw 
a  tangent  at  any  point  equal  in 

length  to  any  given  portion  of      Fra.  117.— The  rectification  of  a 
,  i  curved  line, 

the  curve. 

First. — To  rectify  a  portion  of  any  curve. 

Let  the  curve  be  AB,  Fig.  117.  Set  the  dividers  conveniently 
small,  so  as  to  lay  off  an  arc  whose  chord  is  approximately  the 
length  of  the  arc.  Step  off  with  the  dividers  the  points  1,  2,  3,  4, 
etc.,  from  B  toward  A,  where  the  tangent  is  drawn.  Lay'  off 
with  a  different  pair  of  dividers  the  last  step,  5A  (which  is  not 
likely  to  be  the  same  as  the  others),  on  the  tangent  A5'.  Then, 
with  the  original  dividers,  step  off  from  that  point  the  points 
4',  3',  2',  1'  and  B'.  This  rectification  is  fairly  good,  if  care  is 
taken. 

Second. — To  rectify  a  small  arc  of  a  circle. 

If  the  arc  be  less  than  60°,  a  convenient  method  is  shown  in 
Fig.  118.  Let  AB  be  the  arc,  and  BD  (of  indefinite  length)  be 
the  tangent.  Extend  the  chord  AB  through  B  one-half  its 
length  to  C.  Using  C  as  a  center,  and  CA  as  a  radius,  strike 
the  arc  AD.  BD  will  then  equal  (approximately)  the  arc  BA. 

Third. — To  rectify  a  larger  arc. 

Divide  the  large  arc  into  smaller  ones,  rectify  each,  and  add 
them  together. 


154 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Fourth. — To  rectify  a  quadrant  of  a  circle. 

Let  AB  (Fig.  119)  be  the  given  quadrant.  Draw  the  tangent 
at  B,  and  from  A  draw  a  line  AC,  which  makes  60°  with  CB. 
BC  is  then  a  very  close  approximation  to  AB.  This  is  un- 
doubtedly the  closest  approximation  for  such  an  arc.  Let  the 


FIG.  118. — -The  rectification  of 
a  circular  arc. 


FIG.  119.— The  rectifica- 
tion of  a  quadrant. 


student  prove  by  trigonometry  that  it  is  numerically  equal  to 

7T 

~R.     Although  there  is  no  geometrical  proof  for  this,  your  cal- 

a 

culation  will  show  that  it  is  closer  than  the  accuracy  of  your 
instruments. 

A  convenient  method  of  rectifying  a  small  arc  comes  with 

this  method.  To  rectify  BD, 
Fig.  119,  draw  the  line  ED 
through  to  the  tangent  at  F. 
The  line  BF  equals  BD. 

Fifth.  —  To  rectify  an  arc  of  a 
circle  when  its  angle  is  known. 
Here  the  numerical  method 
is  best.    The  arc  AB  (angle  =  8) 

O  =  .01745R00. 


FIG.  117. 


141.  To  Lay  Off  Any  Length  of  Line,  Straight  or  Curved,  on 
Any  Curve. 

Regard  the  line  AB'  (Fig.  117)  as  a  curved  line.  Lay  off  AB 
on  the  curve  in  exactly  the  same  manner  that  it  was  laid  off  on 
the  straight  line. 


CURVED  LINES  155 

This  is  often  a  necessary  operation  in  practical  drafting; 
example,  in  drawing  cycloids  for  gear  outlines. 

142.  Conic  Sections. 
Definitions. 

Circle. — A  circle  is  a  plane  curve  generated  by  a  point  moving 
so  as  to  be  at  a  constant  distance  from  another  point. 

Ellipse. — An  ellipse  is  a  plane  curve  generated  by  a  point 
moving  so  that  the  sum  of  its  distances  from  two  fixed  points  in 
its  plane  is  constant. 

Hyperbola. — A  hyperbola  is  a  plane  curve  generated  by  a  point 
moving  so  that  the  difference  between  its  distances  from  two 
fixed  points  in  its  plane  is  constant. 

Parabola. — A  parabola  is  a  plane  curve  generated  by  a  point 
moving  so  that  its  distances  from  a  line  and  a  point  in  the  plane 
are  always  equal. 

The  four  foregoing  curves  are  called  " Conic  Sections"  from 
the  fact  that  they  are  the  possible  curves  that  can  be  cut  from  a 
right  circular  cone  by  a  plane.  A  plane  can  also  cut  a  straight 
line  from  a  cone.  The  necessary  conditions  for  cutting  these 
lines  from  a  cone  are: 

Straight  Line. — Plane  through  the  apex. 

Circle. — -Plane  perpendicular  to  the  axis. 

Ellipse. — Plane  oblique  to  the  axis,  cutting  all  the  elements 
on  one  side  of  the  apex. 

Hyperbola. — Plane,  parallel  to  the  axis,  or  a  plane  oblique  to 
the  axis  at  a  smaller  angle  than  that  made  by  the  elements  of  the 
cone. 

Parabola. — Plane  parallel  to  one  element  of  the  cone. 

143.  Gearing  Curves. 

Cycloid. — A  cycloid  is  a  plane  curve  generated  by  a  point  on 
the  circumference  of  a  circle,  as  the  circle  rolls  on  a  straight 
line. 

Epicycloid. — An  epicycloid  is  a  plane  curve  generated  by  a 
point  on  the  circumference  of  a  circle,  as  it  rolls  on  another  circle. 

Hypocycloid. — A  hypocycloid  is  a  plane  curve  generated  by  a 
point  on  the  circumference  of  a  circle  as  it  rolls  on  the  inside  of 
another  circle. 

Involute. — An  involute  is  a  plane  curve  generated  by  a  point 
on  a  tangent,  as  the  tangent  rolls  on  a  plane  curve  or  polygon. 
The  involute  of  a  circle  is  the  only  one  of  any  considerable  im- 
portance, and  is  the  curve  used  in  gearing. 


156  PRACTICAL  DESCRIPTIVE  GEOMETRY 

The  foregoing  are  called  "gearing  curves,"  because  they  are 
mainly  used  as  the  outlines  of  teeth  in  gear  wheels.  The  outlines 
of  bevel  gears  are  (theoretically)  spherical  cycloids,  but  their 
consideration  is  negligible  in  this  work,  and  practically  so  in 
actual  drafting  work. 

SPACE  CURVES 

144.  Helix. — A  helix  is  a  curve  generated  by  a  point  moving 
about  a  straight  line  at  a  constant  distance,  and  in  the  direction 
of  the  line  at  a  constant  velocity. 

Note. — This  line  is  the  line  of  the  screw  thread,  twist  drill,  screw  con- 
veyor, worm  gear,  coiled  spring,  and  many  other  important  practical 
devices.  It  is  a  curve  on  the  surface  of  a  cylinder. 

There  is  also  a  conical  helix  that  sometimes  appears  in  practi- 
cal work.  It  is  usually  drawn  so  that  it  makes  one  complete 
revolution  about  the  axis  in  traveling  from  the  base  to  the  apex, 
ascending  the  surface  at  a  uniform  rate  of  motion.  The  plan  of 
the  conical  helix  is  the  Spiral  of  Archimedes.  By  drawing  the 
plan  first  and  projecting  the  points  of  the  curve  as  it  intersects 
the  various  elements  of  the  cone,  the  elevation  is  easily  drawn. 
It  is  rather  commonly  used  in  sheet  metal  work  and  in  spiral 
springs. 

There  is  no  need  of  defining  or  considering  any  other  of  the 
space  curves. 

GENERAL  DEFINITIONS 

145.  Axis. — The  line  about  which  a  point  revolves,  as  in  the 
case  of  the  helix.     It  is  also  a  center  line  for  some  curves,  as  the 
ellipse,  parabola,  and  hyperbola. 

Minor  Axis  and  Major  Axis. — In  the  ellipse  there  are  two 
dimensions,  sometimes  called  the  long  and  short  diameters,  but 
more  generally  the  major  and  minor  axes.  They  are  respectively 
the  longest  and  shortest  possible  straight  lines  running  through 
the  center  from  one  point  in  the  curve  to  another.  The  constant 
sum,  spoken  of  in  the  definition,  is  equal  to  the  length  of  the 
major  axis. 

Focus. — The  two  fixed  points,  mentioned  in  the  definitions 
of  the  ellipse  and  hyperbola,  and  the  single  one,  mentioned  in 
the  definition  of  the  parabola. 

Vertex. — The  nearest  point  in  the  curve  to  the  focus.  It  has 
the  smallest  radius  of  curvature  of  any  point  in  the  curve,  and 


CURVED  LINES 


157 


might  be  described  as  the  center  of  symmetry  of  the  curve.  The 
circle  has  no  vertex. 

The  constant  difference,  mentioned  in  the  definition  of  the 
hyperbola,  is  the  distance  between  the  two  vertices  of  the  curve. 

Directrix. — The  straight  line  mentioned  in  the  definitions  of 
the  parabola  and  the  cycloid. 

Generating  Circle. — The  rolling  circles  in  the  definitions  of  the 
cycloid,  epicycloid  and  hypocycloid. 

Pitch  Circle,  or  Directing  Circle. — The  circle  on  which  the 
generating  circle  rolls. 

Evolute. — The  curve  on  which  the  tangent  generating  the 
involute  rolls. 

Note. — The  tangent  to  the  evolute  is  normal  to  the  involute. 

Pitch. — The  pitch  of  a  helix  is  the  distance  that  it  travels  in 
the  direction  of  the  axis  in  making  one  complete  revolution. 

PROBLEMS  RELATING  TO  CURVED  LINES 

146.  Problem  47. — To  draw  an  Ellipse,  having  given  its  foci 
and  "constant  sum." 


FIG.  120.— The  ellipse. 

Construction. — 1.  Draw  the  line  AB  (Fig.  120)  and  from  a 
center,  O,  lay  off  F  and  F'  at  equal  distances  from  O,  making 
FF'  equal  to  the  given  distance  between  the  foci. 

2.  On  the  same  line  lay  off  A  and  B  at  equal  distances  from  0, 
making  AB  equal  to  the  given  constant  sum. 

3.  Spread  the  compasses  equal  to  AO,  and  strike  arcs  from 


158 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


F  and  F'  as  centers.     The  intersections,  C  and  D,  of  these  arcs 
are  the  ends  of  the  minor  axis. 

4.  Set  the  compasses  at  arbitrary  radii,  Al,  A2,  etc.,  and 
strike  arcs  with  them  from  F  and  F'  as  centers. 

5.  Set  the  compasses  at  complementary  radii,  Bl,  B2,  etc., 
and  strike  arcs  from  F  and  F'  as  centers. 

6.  Through  A,  B,  C,  D,  and  the  intersections  of  complementary 
radii,  draw  a  smooth  line.     This  will  be  the  required  ellipse. 

Proof. — The  sum  of  the  distances  of  any  of  the  points,  thus 
found,  to  the  foci  is  always  equal  to  AB. 

147.  Approximate  and  Mechanical  Methods  of  Drawing  the 
Ellipse. 


FIG.  121. — Drawing  the  ellipse  with  a  trammel. 

There  are  numerous  ways  of  drawing  an  ellipse,  exact  and 
approximate.  These  are  found  in  most  text-books  on  Me- 
chanical Drawing.  The  best  of  these  is  the  "Trammel  Method." 

The  Trammel  Method. — 1.  Lay  off  the  Major  and  Minor 
Axes,  of  any  desired  dimensions,  AB  and  CD  (Fig.  121). 

2.  Take  a  card,  or  any  straight  edge,  and  lay  off  PI  and  P2, 
respectively  equal  to  the  semi-major  and  semi-minor  axes,  BO 
and  CO. 

3.  Place  the  straight  edge  so  that  2  lies  on  AB  and  1  lies  on 
CD,  and  mark  the  point  P. 

4.  Do  this  in  as  many  positions  as  desired,  marking  a  number 
of  points  on  the  curve. 


CURVED  LINES 


159 


5.  Draw  a  smooth  curve  through  the  points  thus  obtained. 

Uses  for  the  Ellipse. — In  ornamental  work  the  ellipse  is  ex- 
tensively used,  in  its  entirety  and  in  portions.  The  ellipse  is 
the  projection  of  any  circle  in  an  oblique  plane,  and  its  major 
axis  is  always  equal  to  the  diameter  of  the  circle.  In  machine 
design,  the  ellipse  is  used  in  certain  "quick  return"  motions. 
In  sheet  metal  work  and  boiler-making  it  is  often  necessary  to 
cut  elliptical  holes  in  the  plates.  In  designing  "focusing  re- 
reflectors"  the  ellipse  is  necessary,  because  all  rays  from  one 
focus  of  an  ellipse  reflect  from  the  curve  to  the  other  focus. 

Limits. — When  the  distance  between  foci  is  zero,  the  ellipse 
is  a  circle.  When  it  is  equal  to  the  "  constant  sum,"  the  ellipse 
is  a  straight  line. 

148.  Problem  48. — To  draw  the  tangent  to  an  Ellipse  at  a 
given  point  on  the  curve. 


FIG.    120.— The  ellipse. 

Construction. — Let  M  (Fig.  120)  be  the  point. 

1.  Draw  MF  and  MF',  the  focal  lines. 

2.  Extend  MF'  beyond  M  to  F". 

3.  Bisect  the  exterior  focal  angle  FMF". 
The  bisector  MX  will  be  the  tangent  at  M. 

149.  Problem  49. — To  draw  the  tangent  to  an  Ellipse  from  a 
point  outside. 

Let  N  (Fig.  120)  be  the  point. 

1.  Draw  a  circle  with  N  as  center  and  NF'  as  radius. 


160 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  Intersect  the  circle  in  points  P  and  Q  by  an  arc  whose 
radius  is  equal  to  AB,  and  whose  center  is  at  F. 

3.  Connect  either  P  or  Q  with  F. 

4.  Where  PF  or  QF  intersect  the  ellipse  at  R  or  S  will  be  the 
points  to  which  tangents  can  be  drawn  from  N. 

If  the  axes  of  an  ellipse  are  given,  the  foci  may  be  determined 
by  striking  an  arc  from  C  or  D  as  center,  with  a  radius  equal  to 
AO.  Where  this  arc  intersects  AB  will  be  the  foci. 

150.  Problem  50. — To  draw  a  Hyperbola,  having  given  its 
foci  and  the  "  constant  difference." 

Let  F  and  F'  (Fig.  122)  be  the  foci,  and  the  "constant  differ- 
ence" be  equal  to  the  distance  VV. 


FIG.  122.— The  hyperbola. 


Construction. — 1.  Lay  off  F  and  F'  and  V  and  V  on  the  same 
line,  at  their  proper  distances,  all  symmetrically  placed  about  a 
center,  O.  (V  and  V  will  be  the  vertices.) 

2.  Spread  the  compass  at  any  arbitrary  spans,  as  VI,  V2,  etc., 
and  strike  arcs  with  these  radii  with  F  and  F'  as  centers. 

3.  Using  the  complementary  radii,  V'l,  V'2,  etc.,  strike  arcs 
with  F  and  F'  as  centers,  intersecting  those  already  drawn. 

4.  Draw  a  smooth  curve  through  the  proper  intersections  and 
V  and  V7.     This  will  be  the  required  hyperbola. 

Note. — The  hyperbola  is  in  reality  two  curves,  each  having  a  vertex,  and 
each  unlimited  in  extent.  The  hyperbola  varies  in  accordance  with  the 
ratio  between  FF'  and  VV.  The  most  important  use  of  the  hyperbola  is 
that  it  is  the  generating  line  of  the  pitch  surface  of  "skew"  gears. 


CURVED  LINES 


161 


151.  Problem  51. — To  draw  the  tangent  to  a  Hyperbola  from 
any  point  of  the  curve. 

Construction. — Let  P  be  the  point. 

1.  Draw  PF  and  PF',  the  focal  lines. 

2.  Bisect  the  angle  FPF'. 

The  bisector  will  be  the  tangent. 

Note. — The  construction  is  not  shown  in  the  figure.     Let  the  student 
make  it  for  himself. 

152.  Problem  52. — To  draw  a  tangent  to  a  Hyperbola  from 
any  point  outside  the  curve. 

Let  N  (Fig.  122)  be  the  given  point. 

Construction. — 1.  With  N  as  a  center,  draw  a  circle  passing 
through  F',  the  nearest  focus. 


FIG.  123.— The  parabola. 

2.  With  F,  the  farther  focus,  as  the  center,  and  VV  as  the 
length  of  the  radius,  strike  an  arc,  cutting  the  first  circle  in  the 
points  X  and  Y. 

3.  Draw  XF  and  YF,  extending  them  until  they  intersect  the 
hyperbola  in  the  points  P  and  Q. 

4.  NP  and  NQ  will  be  the  required  tangents. 

153.  Problem   53. — To   draw  a  Parabola,   having  given  the 
focus  and  the  directrix. 

Construction. — Let  XD  (Fig.  123)  be  the  directrix  and  F  the 
focus. 
11 


162  PRACTICAL  DESCRIPTIVE  GEOMETRY 

1.  Draw  a  line  CG  through  F  perpendicular  to  XD.     This  is 
the  axis  of  the  curve. 

2.  Bisect  the  distance  from  F  to  XD  at  E.     E  will  be  the 
vertex. 

3.  Draw  a  number  of  lines  at  points,  1,  2,  3,  etc.,  parallel  to 
XD,  at  arbitrary  distances. 

4.  Locate  two  points  on  each  of  the  lines,  equally  distant  from 
F  and  XD. 

5.  Draw  a  smooth  line  through  the  points  thus  found,  and  it 
will  be  the  required  parabola. 

154.  Problem  54. — To  draw  the  Tangent  to  a  Parabola  at  any 
point  on  the  curve. 

Construction. — Let  M  (Fig.  123)  be  the  given  point. 

1.  Draw  MX,  perpendicular  to  XD. 

2.  Draw  the  focal  line  MF. 

3.  Bisect  the  angle  FMX.     The  bisector  is  the  tangent. 

Note. — Extend  the  line  XM  in  the  direction  MY,  and  draw  the  normal 
MO.  Notice  that  MY  is  parallel  to  the  axis,  and  that  MO  is  the  bisector 
of  the  angle  FMY,  making  OMF  equal  to  OMY.  This  fact  is  taken  advan- 
tage of  in  designing  the  reflector  for  the  searchlight.  M  is  a  random  point" 
on  the  parabola,  and  shows  that  any  ray  from  F  striking  the  parabola  at 
M  (any  point)  is  reflected  in  a  line  parallel  to  the  axis,  as  MY.  This  con- 
centrates all  the  reflected  rays  into  a  cylinder  of  light,  preventing  diffusion. 

155.  Problem  55. — To  draw  a  Tangent  to  a  Parabola  from  a 
point  outside. 

Construction. — Let  N  (Fig.  123)  be  the  given  point. 

1.  With  N  as  a  center,  draw  a  circle  through  F. 

2.  From   its  intersections,   C   and   D,    with   DX,    draw   the 
perpendiculars,  CE  and  DK. 

3.  E  and  K,  the  intersections  with  the  curve,  are  the  points 
to  which  tangents  can  be  drawn. 

156.  Problem  56. — To  draw  a  Cycloid,  having  given  the  direc- 
trix and  generating  circle. 

Construction. — Let  AB  be  the  directrix  and  O  be  the  center 
of  the  generating  circle,  which  is  to  be  drawn  tangent  to  the 
directrix,  of  a  radius  R. 

1.  Lay  off  on  AB,  a  distance  equal  to  2-irH,  and  divide  the 
circle  and  AB  into  any  convenient  number  of  equal  parts,  12, 
16  or  24. 

2.  Draw  the  path  of  the  center  of  the  circle,  parallel  to  AB, 
at  a  distance  equal  to  R. 


CURVED  LINES 


163 


3.  Erect  perpendiculars  to  the  line  of  centers  at  each  of  the 
divisions  of  AB.     The  intersections  will  be  successive  positions 
of  the  center  of  the  generating  circle. 

4.  From  the  divisions,  1,  2,  3,  etc.,  of  the  generating  circle, 
run  projectors  parallel  to  AB. 

5.  From  each  of  the  successive  positions  of  the  center,   1', 
2',  3',  etc.,  describe  arcs  intersecting  the  corresponding  projectors. 

6.  Through  these  intersections  draw  a  smooth  curve,  which 
will  be  the  required  Cycloid. 


B 


FIG.  124.— The  cycloid. 


157.  Problem  67. — To  draw  the  tangent  to  the  Cycloid  at  any 
point  on  the  curve. 

Construction. — Let  M  (Fig.  124)  be  the  given  point,  and  P  be 
the  point  of  tangency  with  AB  of  the  generating  circle,  when  M 
is  its  intersection  with  the  curve. 

1.  Draw  MP.     This  line  is  the  normal  at  M. 

2.  Draw  the  tangent  perpendicular  to  MP. 

Note. — It  is  owing  to  the  fact  that  the  normal  to  the  cycloid  at  any  point 
passes  through  the  point  P,  where  the  generating  circle  and  directrix  are 
tangent,  that  the  cycloid  is  a  correct  curve  for  use  in  gear  outlines.  It  is 
this  property  that  makes  it  possible  to  maintain  a  constant  angular  velocity 
ratio  in  gears,  as  will  be  seen  when  the  study  of  Kinematics,  or  "  Mechanism," 
is  taken  up. 

158.  Problem  58. — To  draw  an  Epicycloid  and  a  Hypocycloid. 

The  only  difference  between  drawing  these  curves  and  the 
cycloid,  is  that  the  directrix  now  is  a  circle  instead  of  a  straight 
line.  Fig.  125  shows  both  curves.  AB  is  an  arc  of  the  pitch 
circle,  as  the  directrix  is  usually  called  in  Machine  Design,  and 
the  center  of  the  generating  circle  of  the  Epicycloid  is  at  C,  and 
its  path  is  the  arc  CD  (center  at  O).  The  center  of  the  Hypo- 
cycloidal  circle  is  at  E,  and  its  path  is  the  arc  EF,  center  at  0. 


164  PRACTICAL  DESCRIPTIVE  GEOMETRY 

Construction.— 1.  Divide  the  generating  circles  into  any  con- 
venient number  of  equal  parts. 

2.  Step  off  these  arcs  on  AB  (Art.  141). 

3.  From  the  points  1,  2,  3,  etc.,  draw  arcs  with  their  centers 
at  O. 

4.  Through  the  divisions  on  AB  draw  radial  lines  intersecting 
CD  and  EF  in  the  points  occupied  by  the  centers  as  the  circles 
roll  on  AB. 


FIG.  125. — The  epicycloid  and  hypocycloid. 

5.  From  these  center  points  strike  arcs  with  radii  equal  to 
those  of  the  generating  circles,  intersecting  the  arcs  of  opera- 
tion (3). 

6.  Through  the  respective  intersections  draw  the  curves. 
The  tangents  are  drawn  precisely  as  the  tangents  to  the  cycloid 

are  drawn;  that  is,  the  normal  at  any  point  passes  through  the 
point  of  tangency  of  the  generating  circle  and  the  pitch  circle. 

Note. — The  hypocycloid  generated  by  a  circle  half  the  size  of  the  pitch 
circle  is  a  straight  line  and  is  called  a  radial  hypocycloid. 

159.  Problem  59. — To  draw  an  Involute  to  a  Circle. 

Let  AB  (Fig.  126)  be  the  directing  circle,  or  Evolute,  with  its 
center  at  O. 

Construction. — 1.  Divide  any  arc  of  AB,  as,  for  example,  the 
semicircle,  into  any  number  of  equal  parts. 

2.  At  the  points  1,  2,  3,  etc.,  draw  tangents  in  the  direction  of 
the  origin,  A. 


CURVED  LINES 


165 


3.  Rectify  each  arc  from  the  origin  and  lay  off  the  distance  on 
the  tangent. 

4.  Through  the  extremities  of  the  tangents  draw  the  curve. 

Note  1. — This  curve  is  that  described  by  the  end  of  a  taut  cord  as  it  is 
unwound  from  a  post.  It  is  not  a  closed  curve,  but  a  spiral  that  keeps 
growing  in  expanse  with  every  revolution. 

Note  2.— An  involute  can  be  described  in  the  same  way,  using  any  curve 
or  polygon  as  the  evolute. 

Note  3. — The  Volute,  used  in  architecture,  is  a  spiral,  the  involute  of  an 

enlarging  square. 

^.       i      .^ 

X 


FIG.  126.— The  involute.     , 

160.  Problem  60. — To  draw  the  Tangent  to  an  Involute  at  any 
point  on  the  curve. 

The  tangents  to  the  evolute  are  all  normals  to  the  involute 
(Art.  143),  hence  all  that  is  neces- 
sary is  to  draw  the  perpendicular 
to  the  normal  at  the  given  point. 
Thus  at  the  point  C  (Fig.  126) 
draw  CD  perpendicular  to  C8, 
the  normal. 

Note. — The  fact  that  the  normal  to  the 
involute  is  tangent  to  the  evolute  makes 
the  curve  useful  in  gearing  in  the  same 
way  as  in  the  case  of  cycloidal  curves. 

161.  Problem  61. — To  draw  a 
Spiral  of  Archimedes. 

Construction.— (See  Fig.   127). 

1.  Draw  a  series  of  lines,  radi- 
ating from  O  at  equal  angles.  FlG   127. 

2.  Lay  off  on-  line  1,  one  unit, 

on  line  2,  two  units,  on  line  3,  three  units,  etc. 

3.  Draw  the  curve  through  these  points. 


166 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


162.  Problem  62. — To  draw  the  projections  of  a  Helix,  having 
given  its  Pitch  and  Diameter. 

Let  OP  (Fig.  128)  be  the  axis,  OV12  the  pitch,  and  D  the 
diameter. 

Construction. — 1.  Draw  the  circle  and  divide  it  into  any  num- 
ber of  equal  parts. 

2.  Divide  the  pitch  into  the  same  number  of  equal  parts. 


FIG.  128. 

3.  Draw  perpendiculars  to  the  axis  from  each  of  the  divisions 
made  in  operation  (2). 

4.  Draw  projectors  from  each  of  the  divisions  of  the  circle, 
intersecting  the  corresponding  perpendiculars  from  the  axis. 

5.  Draw  the  curve  through  these  intersections. 
163.  The  Angle  of  the  Helix. 

From  the  definition  (Art.  144),  we  see  that  the  generating 
point  on  the  helix  ascends  a  given  distance  in  traversing  a  given 
arc,  and  that  the  rate  of  ascent  is  constant.  This  means  that 
any  element  of  the  curve  makes  a  certain  angle  with  the  hori- 
zontal (the  axis  of  the  helix  being  taken  in  all  these  problems 


CURVED  LINES 


167 


as  vertical),  and  that  all  the  elements  in  any  one  helix  make 
the  same  angle  with  the  horizontal.  Thus,  in  Fig.  128(a),  if 
AB  is  the  length  of  an  elemental  arc  of  the  circle,  and  BC  is  the 
length  of  an  elemental  division  of  the  pitch,  the  hypothenuse  of 
the  right  triangle  ABC  will  yield  the  inclination  and  the  length 
of  an  element  of  the  helix.  Therefore,  if  a  right  triangle,  having 
its  base  angle  equal  to  a,  be  wrapped  about  a  cylinder,  the  hy- 
pothenuse will  become  the  helix.  From  this  it  is  evident  that 

pitch 

tan  a  =  — -p • 

TT.  diam. 

164.  The  Tangent  to  the  Helix. 

If  the  right  triangle,  just  mentioned,  be  unwrapped  from  the 
cylinder,  its  hypothenuse  becomes  the  tangent  to  the  helix 
at  the  point  where  the  unwrapping  ceases. 


FIG.  129. — The  generation  of  a  helix. 


This  is  so  because  the  tangent  must  make  the  same  angle 
with  the  horizontal  that  the  helix  does,  because  it  contains  two 
consecutive  points  (one  element)  of  the  curve. 

Therefore,  the  tangent  becomes  the  helix  if  it  is  rolled  on  the 
cylinder,  and  the  length  of  the  helix  from  its  origin  to  the  point  of 
tangency  is  equal  to  the  length  of  the  tangent  from  the  plane  of 
the  origin  to  the  point  of  tangency.  Referring  to  the  picture, 
Fig.  129,  will  show  that  AO  =  BO,  and  further  that  AP  =  BP. 
AP  is  the  //"-projection  of  the  tangent,  and  BP  is  the  //"-pro- 
jection of  the  helix.  From  this  the  rule  is  derived:  The  H- 
projection  of  the  Helix  is  of  the  same  length  as  the  H-projection 
of  the  Tangent  from  the  plane  of  the  origin  to  the  point  of 
tangency. 


168  PRACTICAL  DESCRIPTIVE  GEOMETRY 

165.  Problem  63. — To  draw  the  Tangent  to  the  Helix  at  any 
point  on  the  curve. 

Construction. — Let  B  (Fig.  128)  be  the  given  point.     1.  Draw 
the  tangent  to  the  circle  at  bH. 

2.  Rectify  the  arc  bH12,  and  lay  it  off  on  the  tangent  aHbH. 

3.  Project  aH  to  av  on  GL. 

4.  avbv  will  be  the  F-projection,  and  aHbH  the  //-projection  of 
the  tangent  required. 

166.  EXERCISES  IN  CURVED  LINES 

Note. — No  ground  line  is  necessary  in  drawing  any  of  the  curves  but  the 
helix. 

745.  Draw  an  ellipse,  AB  =  5  in.,  FF'  =  3  in.     Draw  a  tangent  at  a  point  on 
the  curve  2  in.  from  F.     Draw  two  tangents  from  a  point  4  in.  from  F 
and  2  in.  from  F'. 

746.  Draw  an  ellipse,  AB  =  4^  in.,  and  CD  =  3  in.    Locate  F  and  F',  and 
draw  a  tangent  at  a  point  on  the  curve  1  in.  from  A.     Draw  two 
tangents  from  E,  which  is  3  in.  from  F'  and  2  in.  from  F. 

747.  A  and  B  are  two  tangent  ellipses  of  equal  size.     The  foci  of  A  are 
C(l,  0-2)  D(3^,  0  -  2),  and  the  more  distant  focus  of  B  is  at  E 
(5,  0  +  1).     Draw  the  two  ellipses. 

748.  Draw  a  hyperbola.     FF'  =  3  in.  and  the  "constant  difference"  =2  in. 
Draw  the  tangent  at  a  point  on  the  curve  2  in.  from  F.     Draw  a 
tangent  from  a  point  2  in.  from  F'  and  2^  in.  from  F. 

749.  Draw  a  parabola  with  the  distance  between  focus  and  directrix  1  in. 
Draw  tangents  from  assumed  points  on  and  outside  the  curve. 

750.  Draw  a  parabola  whose  distance  from  focus  to  vertex  is  %  in.     Draw 
tangents  from  assumed  points  on  and  outside  the  curve. 

751.  Draw  a  cycloid  with  a  IJ^-in.  generating  circle.     Draw  a  tangent  at 
the  point  on  the  curve  at  fie  of  the  revolution. 

752.  Draw  an  epicycloid  and  a  hypocycloid  on  a  pitch  circle  of  6-in.  radius, 
with  1^-in.  generating  circles.     Draw  the  normals  to  each  curve  at 
fie  of  a  revolution  from  the  origin. 

753.  Draw  an  involute  of  a  1-in.  circle. 

754.  Draw  an  involute  of  a  J^-in.  square. 

755.  Draw  an  involute  of  a  hexagon  of  J^-in.  sides. 

756.  Draw  an  involute  of  a  %-in.  equilateral  triangle. 

757.  Draw  a  helix  of  2-in.  diameter,  3-in.  pitch.     Draw  its  tangent  at  60° 
from  the  origin.     Also  one  at  90°. 

758.  Design  a  focusing  reflector  of  diameter  —   —  in.  (4,  5,  6,  7),  in  which 
the  rays  from  a  light,  placed  -    —  in.  (2,  3,  4,  5)  from  the  vertex  are 
concentrated  at  a  point in.  (8,  9,  10,  12,  14,  16)  from  the  light. 


CHAPTER  VI 
SINGLE-CURVED  SURFACES 

167.  Definitions. 

A  line,  straight  or  curved,  moving  in  accordance  with  some 
law,  generates  a  Surface.  If  the  generating  line  be  straight,  and 
moves  along  two  parallel  or  intersecting  straight  lines,  the 
surface  generated  will  be  a  Plane.  There  is  but  one  kind  of 
plane,  all  other  kinds  of  surfaces  are  curved. 

Curved  Surfaces  are  divided  into  three  classes:  Single- 
Curved,  Double -Curved  and  Warped  surfaces. 

A  Single -Curved  Surface  is  generated  by  a  straight  line  follow- 
ing a  curved  director  in  such  a  way  that  any  two  consecutive 
positions  of  the  line  shall  lie  in  a  plane. 

A  Warped  Surface  is  generated  by  a  straight  line  moving  so 
that  no  two  of  its  consecutive  positions  shall  be  in  the  same  plane. 

Note. — Single-curved  and  warped  surfaces  are  sometimes  called  Ruled 
surfaces,  because  they  are  made  up  of  straight  lines.  Through  every  point 
of  a  ruled  surface  at  least  one  straight  line  can  be  drawn  in  the  surface. 

A  Double -Curved  Surface  is  generated  by  a  curve  moving 
along  a  curved  director,  so  that  there  are  no  straight  lines  in  it. 

Generatrix. — The  line  generating  a  surface. 

Element. — Any  position  of  the  generatrix. 

Consecutive  Elements. — Two  successive  positions  of  the 
generatrix,  having  no  assignable  distance  between  them. 

A  Surface  of  Revolution  is  a  surface  generated  by  the  revolu- 
tion of  any  line,  straight  or  curved,  about  a  straight  line  axis. 
Surfaces  of  revolution  may  be  single-curved,  double-curved,  or 
warped. 

168.  Classification  of  Surfaces. 
Single-Curved  Surfaces  are  of  three  varieties: 
Cones,  all  the  elements  intersect  in  a  point. 
Cylinders,  all  the  elements  are  parallel. 

Convolutes,  consecutive  elements  intersect  two  and  two,  no 
three  elements  having  a  common  point. 

Double-Curved  Surfaces  are  mostly  surfaces  of  revolution, 
spheres,  ellipsoids,  paraboloids,  tori,  and  combination  surfaces. 

169 


170 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


They  are  sometimes  convex,  sometimes  concave,  and  sometimes 
both.  The  annular  torus  has  both  in  continuous  surface. 

Warped  Surfaces  are  numerous  in  their  variety,  but  the  most 
important  are  the  helicoid  and  hyperboloid  of  revolution  of  one 
nappe. 

169.  The  following  table  will  give  a  convenient  summary  of 
the  surfaces. 


Single- 
Curved 


Warped 


Double- 
Curved 


Surfaces  of 
Revolution 


Cone — all  elements  intersecting 

Cylinder — all  elements  parallel 

Convolute — consecutive    elements    intersect- 
ing. 

Helicoid 

Hyperboloid  of  Revolution  r  Ruled  Surfaces. 

Hyperbolic  Paraboloid 

Conoid 

Cylindroid 

Warped  Cone 

Cow's  Horn. 

Sphere 

Oblate  Spheroid 

Prolate  Spheroid 

Paraboloid  (both  varieties) 

Hyperboloid  of  Revolution  of  Two  Nappes 

Torus. 

f  Right  Circular  Cone,  Right  Circular  Cylinder 
I  Hyperboloid  of  Revolution  of  One  Nappe 
[  Sphere,  Spheroids,  Ovoids,  etc. 


170.  Tangent  Planes  and  Developments. 

The  most  important  problems  in  surfaces  are  to  be  found  in 
drawing  tangent  planes  to  surfaces,  the  intersections  of  surfaces, 
and  developing  the  surfaces. 

A  Tangent  Plane  to  a  surface  at  any  point  is  a  plane  composed 
of  all  the  tangent  lines  to  all  the  lines  of  a  surface  passing  through 
that  point.  As  two  intersecting  lines  determine  a  plane,  the 
general  rule  for  tangent  planes  is  this:  Any  two  tangent  lines  at 
any  point  on  any  surface  determine  the  tangent  plane  at  that 
point. 

Any  plane,  passed  through  the  point  of  tangency  so  as  to 
intersect  a  surface  and  its  tangent  plane,  will  cut  from  them  a 
line  and  its  tangent. 

In  ruled  surfaces,  one  and  sometimes  two  straight  lines  of 
the  surface  can  be  drawn  through  each  point  in  the  surface, 
therefore  this  line  (element,  as  it  always  is)  can  be  used  for  one 


SINGLE-CURVED  SURFACES 


171 


of  the  tangents  in  the  foregoing  rule.  This  is  true,  because  a 
straight  line  is  its  own  tangent.  On  this  account,  the  rule  will 
read:  "In  ruled  surfaces,  one  tangent  and  the  element  at  any 
point  will  determine  the  tangent  plane  at  that  point." 

Normal. — The  normal  line  at  any  point  of  a  curved  surface  is 
the  perpendicular  to  the  tangent  plane  at  that  point.  The 
normal  plane  is  any  plane  containing  the  normal  line. 

Development. — By  Development  is  meant  the  rolling  out  of  a 
surface  into  a  plane  figure,  which  can  be  rolled,  or  folded,  back 
to  make  the  original  surface.  Only  single-curved  surfaces  are 
strictly  developable,  because  they  only  have  consecutive  elements 


FIG.  130. — Tangent  plane  to  a  cylinder  determined  by  two  tangent  lines. 

lying  in  planes.  Approximate  developments  can  be  made  of 
other  surfaces,  as  will  appear  when  the  surfaces  are  taken  up 
separately. 

THE  CONE 

171.  Definitions. — A  Cone  is  generated  by  a  straight  line 
moving  so  as  to  touch  constantly  a  curved  line  directrix,  while 
one  point  in  the  generatrix  remains  stationary. 

Apex  or  Vertex,  the  stationary  point  on  the  generatrix. 

Base,  a  limiting  plane  that  cuts  all  the  elements. 

Axis,  a  straight  line  through  the  apex  and  the  center  of  the 
base. 


172  PRACTICAL  DESCRIPTIVE  GEOMETRY 

A  Frustum  is  a  portion  of  the  surface  between  two  parallel 
planes  on  the  same  side  of  the  apex,  which  cut  all  the  elements. 

Discussion. — The  surface  of  the  cone  that  lies  between  the 
base  and  apex  is  usually  all  that  is  considered.  Theoretically, 
however,  the  surface  is  unlimited,  and  extends  beyond  both 
limits  to  infinity.  This  theoretical  extension  creates  two 
divisions  of  the  surfaces,  one  on  each  side  of  the  apex,  called 
respectively  the  Upper  Nappe  and  Lower  Nappe.  In  practical 
work  this  condition  almost  never  exists,  and  is  usually  ignored. 

Varieties  of  Cones. — The  different  varieties  of  cones  are  dis- 
tinguished by  their  bases  and  axes.  The  bases  may  be  circular, 
elliptical,  hyperbolic,  etc.,  or  combinations  of  curves,  and  the 
axes  are  either  perpendicular  (right)  or  oblique  to  the  bases. 
Thus  we  designate  a  cone  as  a  right  circular  cone,  an  oblique 
elliptical  cone,  etc.  The  base  and  apex,  or  the  base  and  axis,  are 
sufficient  to  determine  the  cone. 

Limits. — A  cone,  whose  apex  is  at  infinity,  becomes  a  cylinder, 
and  one,  whose  apex  is  in  the  plane  of  the  base,  becomes  a 
plane. 

A  Tangent  Plane  to  a  cone  at  any  point  on  the  surface  is 
tangent  all  along  the  element  through  that  point.  This  element 
is  called  the  Element  of  Contact.  As  all  elements  pass  through 
the  apex,  every  tangent  plane  must  contain  the  apex.  This 
fact  is  of  great  importance  in  passing  tangent  planes.  It  means 
that  any  line  through  the  apex,  except  lines  inside  the  cone, 
will  lie  in  one  or  two  tangent  planes.  From  this  fact,  and  from 
Art.  155,  we  derive  the  special  rule  for  passing  tangent  planes  to 
cones. 

Rule. — One  tangent  line  and  the  apex  are  sufficient  to  deter- 
mine a  tangent  plane  to  any  cone. 

Axiom. — If  the  base  of  a  cone  is  in  H,  the  H-trace  of  any 
tangent  plane  will  be  tangent  to  the  base. 

172.  Development  of  Cones. — When  the  surface  of  a  cone  is 
rolled  out,  we  find  that  there  is  one  point  common  to  all  elements, 
the  apex,  and  this  point  will  remain  a  point  in  the  development. 
From  this  point  all  the  elements  radiate  in  fan  shape,  making 
up  an  integral  of  all  the  minute  triangles  that  form  the  surface 
of  the  cone.  In  the  case  of  a  right  circular  cone,  the  elements 
are  all  of  equal  length,  and  the  base  rolls  out  into  an  arc  of  a 
circle,  whose  radius  is  the  length  of  the  elements  (sometimes 
called  "slant  height").  The  length  of  the  arc  thus  derived  is 


SINGLE-CURVED  SURFACES 


173 


equal  to  the  circumference  of  the  base.  All  other  cones  have  ele- 
ments of  varying  length,  and  their  bases  will  not  roll  out  in  cir- 
cular paths.  The  development  of  all  cones,  except  the  right 
circular,  is  accomplished  by  "triangulation ;"  that  is,  dividing 


FIG.  131. — A  cone  rolled  out  in  a  plane. 

the  surface  into  small  triangles,  and  laying  them  out  flat  in 
order  and  (approximately)  true  size.  An  exception  to  this  may 
be  found  in  an  oblique  elliptical  cone  that  may  have  a  circular 
right  section.  If  such  be  found,  the  circular  method  is  better 
than  triangulation. 

173.  Problem  64. — To  locate  a  point 
on  the  surface  of  a  cone. 

Construction. — Let  the  F-projection 
of  a  point  (ov  in  Fig.  132)  be  given. 

1.  Draw  the  element  of  the  surface 
through  the  point. 

2.  Project  ov  to  OH  on  the  //"-projec- 
tion of  the  element. 

Note. — Two   points   are  possible  in  this 
case,  unless  front  or  rear  surface  is  specified. 

174.  Problem  65. — To  obtain  the 
line  of  intersection  of  a  Plane  and  a 
Cone. 

Construction. — Exactly  the  same  as 
for  the  intersection  of  a  plane  and    FlG> 
pyramid,  taking   a  reasonable  num- 
ber of  elements  of  the  cone,  instead  of  edges. 

175.  Problem  66. — To  obtain  the  piercing  points  of  a  Cone  by 
a  given  Line. 

Analysis. — 1.  Pass  a  plane   through  the   given  line  and  the 
apex  of  the  cone. 


132. — A  point  on   the 
surface  of  a  cone. 


174 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  Obtain  the  elements  of  the  cone  cut  by  this  plane. 

3.  Where   these   elements   intersect   the   given   line   will   be 
located  its  piercing  points  with  the  cone. 

Let  the  student  make  the  construction. 

Note. — There  are  occasions  when  a  plane  intersecting  a  number  of  ele- 
ments can  be  used  to  better  advantage  than  a  plane  through  the  apex.  Such 
a  plane  will  cut  a  curve  from  the  cone  surface,  and  the  piercing  points  will 
be  at  the  intersections  of  the  given  line  with  the  curve. 

176.  Problem  66. — To  pass  a  plane  Tangent  to  a  Cone  at  a 
point  on  the  surface. 

Analysis. — Read  the  rule  given  in  Art.  171. 

Construction. — Refer  to  Fig.  133.    Let  O  be  the  given  point. 

1.  Draw  the  element  through   O. 
This  is  the  Element  of  Contact,  and 
one  line  of  the  plane. 

2.  Where  the  element  of  contact 
intersects  the  base,  draw  the  tangent 
to  the  base. 

This  is  the  //"-trace  of  the  required 
plane,  and  the  F-trace  may  be  ob- 
tained in  several  ways  known  to  the 
student. 

177.  Problem  67. — To  pass  a  plane 
Tangent  to  a  Cone  through  a  Point 
outside  the  surface. 

Analysis. — 1.  Draw  a  line  through 
the  apex  and  the  given  point.  (See 
Art.  171.) 

2.  Find  the  piercing  point  of  this 
line  with  the  plane  of  the  base. 

3.  From  this  point  draw  one  of  the  two  possible  tangents  to 
the  base. 

4.  Pass  the  required  plane  through  the  tangent  line  and  the 
apex. 

Construction. — Let  the  student  make  it. 

Note  1. — If  the  line  through  the  given  point  and  the  apex  is  parallel  to 
the  base,  the  tangent  line  in  the  plane  of  the  base  will  be  parallel  to  said 
line,  and  may  be  drawn  on  either  side  of  the  base. 

Note  2. — If  the  line  through  the  apex  is  too  nearly  parallel  to  the  plane  of 
the  base  to  obtain  the  piercing  point  within  the  limits  of  the  drawing,  a 
plane  may  be  passed  parallel  to  the  base,  nearer  the  apex,  and  a  tangent 


FIG.  133. 


SINGLE-CURVED  SURFACES 


175 


may  be  drawn  to  the  new  base,  or  a  second  cone  similar  to  the  first  may  be 
drawn  with  its  apex  in  the  line  at  another  point,  and  the  tangent  plane 
drawn  to  both  cones. 

178.  Problem  68. — To  pass  a  plane  Tangent  to  a  Cone  parallel 
to  a  given  line. 

Analysis. — 1.  Draw  a  line  through  the  apex  parallel  to  the 
given  line. 

2.  Having  this  line,  finish  as  in  Problem  67. 

Construction. — Let  the  student  make  it. 

Impossible  Conditions. — When  the  line  through  the  apex 
parallel  to  the  given  line  is  inside  the  cone,  the  problem  is  impos- 
sible. In  such  a  case  the  line  will  pierce  the  plane  of  the  base 
inside  the  base.  Otherwise  the  problem  is  possible. 

179.  Problem  69. — To  pass  a  plane  Tangent  to  a  Cone,  making 
a  given  angle  with  H. 


Limitations. — Only  certain  conditions  will  allow  a  solution  of 
this  problem. 

1.  If  the  cone  be  a  right  circular  cone,  with  its  base  on  H, 
there  is  only  one  angle  possible,  and  that  is  the  angle  that  its 
elements  make  with  H. 

2.  If  the  base  be  in  some  plane  not  parallel  to  H,  there  are 
many  angles  at  which  tangent  planes  may  be  passed,  though  not 
all. 


176 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


3.  If  the  base  be  in  H,  and  the  axis  oblique,  different  angles, 
but  not  all,  may  be  assigned  to  the  problem. 

4.  If  the  base  be  in  H,  and  a  different  curve  from  the  circle, 
even  though  the  axis  be  perpendicular,  there  will  be  some  variety 
to  the  angles  that  may  be  required. 

Each  problem  will  fix  its  own  limitations. 
Construction. — Let  it  be  required  to  pass  a  plane  tangent  to 
the  cone  AB  (Fig.  134)  making  60°  with  H. 


FIG.  135. 

1.  With  B  as  its  apex,  draw  the  projections  of  a  60°  cone, 
axis  BC. 

2.  Draw  the  //-trace  of  T,  tangent  to  the  bases  of  the  two 
cones.     This  may  be  drawn  on  either  side  of  the  cones. 

3.  Locate  the  F-trace  of  T  by  any  lines  through  B,  lying  in  T. 

Note. — Under  certain  conditions  four  planes  can  be  passed  making  the 
same  angle  with  H,  all  tangent  to  the  cone. 

180.  Problem  70. — To  develop  the  surface  of  a  right  circular 
cone. 


SINGLE-CURVED  SURFACES 


177 


Analysis. — Read  Art.  172. 

Construction. — Let  it  be  required  to  develop  the  truncated 
cone  in  Fig.  135;  that  is,  the  surface  included  between  the  base 
and  the  plane  T. 

1.  Divide  the  base  into  any  convenient  number  of  equal  parts, 
and  draw  the  elements. 

2.  From  bv,  the    apex,  on  the  F-projection,  describe  an  arc  of 
a  radius  to  the  true  length  of  the  elements. 

3.  Make  this  arc  equal  in  length  to  the  circumference  of  the 
base  of  the  cone,  and  divide  it  into  the  same  number  of  equal 
parts.     (See  Note  below.) 


Deve\opnnenT. 


FIG.  136. 


4.  Measure  the  true  length  of  the  truncated  elements,  and  lay 
them  off  on  the  corresponding  radii  of  the  Pattern  (development). 

5.  Draw  a  smooth  curve  through  these  points. 

Note. — The  accurate  way  to  lay  off  this  arc  equal  in  length  to  the  circum- 

T>  ^fiO°  T 

ference  is  by  this  equation:  —  =  —^r,  or  6°  =  ^360°,  where  R  equals  the 

r          v  j\ 

slant  height  of  the  cone,  r  the  radius  of  the  base,  and  0  is  the  number  of 
degrees  in  the  angle  of  the  development.  It  can  be  done  more  roughly  by 
stepping  off  small  arcs  with  the  dividers. 

181.  Problem  71. — To  develop  the  surface  of  an  oblique  cone. 

This  is  accomplished  by  triangulation,  and  it  is  as  nearly  exact 
as  the  chord  of  a  small  arc  is  to  the  length  of  the  arc. 
Construction,— The  cone  PO  (Fig.  136). 
12 


178 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


1.  Divide  the  base  into  a  number  of  equal  arcs. 

2.  Draw  the  elements  of  these  divisions,  dividing  the  cone 
into  small  (approximate)  triangles. 

3.  Measure  the  true  length  of  all  the  elements. 

Note. — By  a  little  care  in  the  division  this  can  be  done  in  pairs. 

4.  Rectify  one  of  the  equal  arcs. 

5.  Beginning    (preferably)    with    the    longest    element,    01, 
construct  triangles  equal  to  the  (approximate)  true  size  of  those 
in  the  cone,  as  shown  in  the  development  at  the  right. 


FIG.  137. — Reversing  the  upper  part  of  the  cone  to  make  an  elbow. 


182.  The  Problem  of  the  Shortest  Path. 

The  shortest  line  that  can  be  drawn  on  any  surface  from  one 
point  to  another  is  a  straight  line  on  the  development.  Let 
A  and  B  (Fig.  136)  be  two  points  on  the  surface  of  the  cone. 

1.  Locate  them  on  the  development,  see  ai  and  bi. 

2.  Draw  a  straight  line  from  ai  to  bi,  and  locate  its  intersec- 
tions, Ci,  di,  etc.,  with  the  intervening  elements. 

3.  Return  all  these  points  to  their  original  positions  in  the 
projections  of  the  cone,  and  draw  the  curve  through  them. 


CONICAL  ELBOWS 

183.  Principle. — A  plane  section,  through  all  the  elements 
of  a  cone,  as  in  Fig.  137,  is  an  ellipse,  which  is  a  symmetrical 
curve.  A  symmetrical  curve  can  be  inverted  upon  itself. 


SINGLE-CURVED  SURFACES 


179 


Construction. — 1.  If  it  is  desired  to  have  a  conical  elbow  of 
two  pieces  to  turn  through  an  axial  angle  of  26,  pass  a  plane 
through  the  cone  at  an  angle  of  9  with  the  base. 

2.  Turn  the  truncated  portion  on  the  lower  part,  and  the 
required  elbow  is  made. 


\i 


FIG.  138. — Center  line  layout. 


MULTI-JOINT  ELBOWS 

183.  The  construction  of  this  is  essentially  that  of  the  single 
joint  elbow. 

D 


FIG.  139. — Laying  out  the  seams. 

Example. — Required  to  reduce  a  6-in.  pipe  to  a  2-in.  pipe, 
through  a  90°  bend,  in  three  pieces. 

1.  Layout  the  center  lines  according  to  Fig.  138. 


180 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  Lay  off  the  total  length  of  the  center  lines  as  the  altitude 
of  the  frustum  of  a  cone  of  6-in.  base  and  2-in.  top. 

3.  Cut  the  frustum  by  planes  through  B  and  C  at  the  angles 
shown  in  Fig.  139. 

4.  Reverse  these  truncated  portions  as  shown  in  Fig.  140. 


184. 


FIG.  140. — Three-piece  elbow. 
EXERCISES 


aV 


6  Cone  Tangent 
toH       ,  v 

D 


Cone  Tangent 
toH      ,v 


10 


Cone  Tanqent 
toV 


Graphic  Layout  No.  60.  —  Center  lines  of  right  circular  cones. 
Base  --  in.  (1,  1^,  IK,  2,  2^). 
Altitude  --  in.  (\Y2,  2,  2)4,  3,  3^). 

Note.  —  All  these  axes  are  in  7.     The  signs  may  be  reversed,  and  the  lines 
drawn  in  ///  at  the  option  of  the  instructor. 

Dimensioned  Layout.  —  Center  lines.     Let  A  or  B  be  the  center  of  the 
base. 


(1)  A(3+3  - 

(2)  A(l  +  1  -  1H) 


,  0  - 

+  1  -  IK). 


SINGLE-CURVED  SURFACES 


181 


<3)  A(2  +  1  -  1)  B(4  +  1  -  2). 

(4)  A(2  +  1  -  IK)  B(4  +  2K  -  IK). 

(5)  A(3  +  2K  -  1)  B(3  +  1  -  2). 

(6)  A(3  +  2K  -  1)  B(3  +  1  -  2). 

(7)  A(2,  0  -  1)  B(4  +  1  -  1).     (Cone  tangent  to  77.) 

(8)  A(2,  0  -  1)  B(4  +  1  -  2).     (Cone  tangent  to  //.) 

(9)  A(2  +  1  -  1)  B(4  +  2  -  2K). 

(10)  A(3  +  IK,  0)  B(3  +  IK  -  3). 

(11)  A(2  +  1,  0)  B(4  +  1  -  IK)-     (Cone  tangent  to  V.) 

Note. — The  above  center  lines  may  be  drawn  in  777,  reversing  the  signs 
+  and  — ,  if  preferred. 


j 


VM 


HM' 


H*N 
VN 


/ 


t 


Graphic  Layout  No.  50A. — Cutting  planes.     (Note. — The  signs  on  these 
planes  may  be  reversed,  at  the  option  of  the  instructor.) 


1     c' 


Graphic  Layout  No.  51. — Piercing  lines.  (Note. — If  the  cone  is  drawn  in 
777,  these  lines  should  be  drawn  in  777.  Draw  them  so  that  they  are  likely 
to  pierce  the  cone.) 

Cutting  Planes. — Traces  may  be  reversed,  if  desired. 

(1)  Q(1K  +3)  1K(5  -2). 

(2)  8(5  +  3)  1(5  -  2). 

(3)  T(l  +  3)  2(5  -  2). 

(4)  U(+2)»(  -2). 


182 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


(5)  W(-l)»(  -M). 

(6)  X(l  +  3)  3(5  -  3). 

(7)  Y(l  +  1)  5(1  +  2M). 

(8)  Z(2  +  3)  3(2  -  2). 

(9)  P(3M  +  3)  3y2(3y2  -  3). 

Piercing  Lines.  —  Signs  may  be  reversed,  if  desired. 


(1)  A(2,  0  -  1M)  B(4  +  2  -  2). 

(2)  C(2,  0  -  2)  B(4  +  2  -  M). 

(3)  E(3  +  2  -  M)  F(3  +  1  - 

(4)  G(3  +  2  -  2)  K(3  +  M  -  1). 

(5)  M(l  +  1  -  1M)  N(4  + 


759.  Locate  a  point  on  the 

—  in.  away  from  H. 

760.  Cut  the  surface  of  cone 
the  section. 


(front   or  rear)  surface  of  cone   , 

by  plane .     Find  the  true  size  of 


Graphic  Layout  No.  62. — Oblique  cones  with  circular  bases,  diameters 

1  J«5  in.,  1%  in.,  2  in.,  23^  in.,  and  axes  1^  in.,  2  in.  23^  in.,  3  in.  long. 

Oblique  cones  with  elliptical  bases,  1  in.  X  1M  m.,  1  in.  X  2  in.,  1^  in.  X 

2  in.,  11A  in.  X  2^  in.,  2  in.  X  2^  in.,  and  axes  1%,  2,  2^,  3  in.  long. 
Use  this  Layout  in  Ex.  770-781. 

761.  Develop  the  surface  between  the  base  and  cutting  plane,  in  Ex.  760. 

762.  Develop  the  surface  between  the  apex  and  cutting  plane,  in  Ex.  760. 

763.  Find  the  points  in  which  the  surface  of  cone  is  pierced  by  line 


between  the  piercing 


764.  Find  the  shortest  line  on  the  surface  of  cone 
points  of  line . 

765.  Assume  a  point  on  the  surface  of  cone ,  and  draw  the  traces  of  a 

plane  tangent  at  that  point. 

766.  Assume  a  point  in  space  outside  the  cone,  and  draw  the  traces  of  a 
plane  (or  two  planes,  if  desired)  tangent  to  the  cone . 

767.  Draw  the  traces  of  one  of  the  tangent  planes  to  cone ,  inclined 

°  to (H,  V,  or  P). 


SINGLE-CURVED   SURFACES  183 

768.  Draw  the  traces  of  the  tangent  plane   (or  planes)  to  the  cone  , 

parallel  to  line  —  —  (one  of  the  given  piercing  lines). 

769.  What  are  the  limitations  of  the  specifications  of  Ex.  766  and  767,  if 

any? 

Dimensioned  Layout. — Let  the  axes  be  the  following  lines,  with  the  cir- 
cular or  elliptical  base  —  —  at  A.     Use  in  Ex.  770-781. 

(1)  A  (2,  0  -  IK)  B(4  +  2  -  IK). 

(2)  A(2,  0  -  1)  B(4  +  2  -  2). 

(3)  A(3  +  1,  0)  B(3  +  2K  -  IK). 

(4)  A(2  +  IK,  0)  B(3K  +  IK  -  2). 

(5)  A(2  +  1,  0)  B(3K  +2-2). 

(6)  A(4  +  1  -  IK)  B(2  +  2  -  IK)- 

(7)  A(4  +  2  -  2)  B(2  +  1  -  K). 

(8)  A(3,  0  -  1)  B(3  +  2  -  2). 

770.  Cone  ,  base  ,  axis  —  — ;  locate  a  point  on  the  (front 

or  rear)  surface, in.  away  from  H. 

771.  Cone  ,  base  ,  axis ;  draw  the  intersection,  and  its  true 

size,  made  by  plane . 

772.  Develop  the  surface  of  the  base  frustum  in  Ex.  770. 

773.  Develop  the  surface  of  the  apex  frustum  in  Ex.  770. 

774.  Cone  ,  base  ,  axis  .     Find  the  points  in  which  it  is 

pierced  by  line . 

775.  Draw  the  shortest  line  on  the  surface  of  the  cone  between  the  pierc- 
ing points  in  Ex.  774. 

776.  Cone  ,  base ,  axis ;  assume  a  point  on  the  surface,  and 

draw  the  traces  of  a  plane  tangent  at  that  point. 

777.  Assume  a  point   in    space,  outside   of    cone ,  base  ,    axis 

;  draw  the  traces  of  a  plane  (or  both  planes,  if  desired)  tangent 

to  the  cone  through  the  assumed  point. 

778.  Cone ,  base ,  axis ;  draw  the  traces  of  a  plane  tangent, 

and  inclined °  to (H,  V,  or  P). 

779.  Cone  —  — ,  base ,  axis  ;  draw  the  traces  of  a  plane  tangent 

to  the  cone  and  parallel  to  line . 

780.  What  are  the  limitations  of  the  specifications  of  Ex.  778  and  779,  if 
any? 

781.  Cone ,  base ,  axis ;  draw  the  projections  and  true  size 

of  its  right  section  at  the  middle  point  of  its  axis. 

782.  Lay  out  the  patterns  for  Funnel  No. .    Graphic  Layout  No.  53. 

783.  Lay  out  the  patterns  for  Motor  Boat  Funnel  No. . 

784.  Lay  out  the  patterns  for  Ventilator  Hood  No. . 

785.  Lay  out  the  patterns  for  Sink  Drainer  No.  . 

786.  Lay  out  the  patterns  for  Offset  Reducing  Pipe  No.  ,  C  =  , 


787.  Lay  out  the  patterns  for  the  Furnace  Hood  No. .     Make  the  plan 

of  the  conical  helix  a  Spiral  of  Archimedes,  making  one  complete 
revolution.     Omit  lap  and  rivet  holes. 

788.  Lay  out  the  patterns  for  Chimney  Transition  Piece  No. .     Graphic 

Layout  No.  54. 


184 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


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Graphic  Layout  No.  63. — Dimensions  in  tables  are  in  inches. 


SINGLE-CURVED   SURFACES 


789.  Lay  out  the  patterns  for  Conical  Elbow  No.  — * — ,  E  = 
37H,  45). 

790.  Lay  out  the  patterns  for  Multi-joint  Conical  Elbow  No. ,  C  = 

°  (75,  90,  105,  120,  135),  number  of  pieces  = (3,  4,  5,  6). 


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Multi-Joini  Conical  Elbow 


Multi-Joint  Oblong  Comcal  Elbow 


Graphic  Layout  No.  54. — Dimensions  given  in  inches,  angles  in  degrees. 


791.  Lay  out  the  patterns  for  Multi-joint  Oblong  Conical  Elbow  No. 

D  =  -°    (75,    90,     105,     120,     135),    number    of    pieces   = 

(3,  4,  5,  6). 
Note. — The  dimensions  for  A  and  C  in  Ex.  791  may  be  transposed. 


186  PRACTICAL  DESCRIPTIVE  GEOMETRY 

THE  CYLINDER 

185.  Definition. — A  cylinder  is  a  single-curved  surface  gene- 
rated by  a  line  moving  along  any  curved-line  directrix,  always 
remaining  parallel  to  a  given  line.     It  is  a  cone  with  its  apex  at 
infinity. 

The  Surface  of  a  cylinder  is  usually  taken  to  be  that  portion 
which  is  included  between  two  plane  bases.  The'  bases  may  be 
parallel  or  not.  Theoretically  the  cylinder  is  infinite  in  extent. 

Axis. — The  axis  is  a  line  through  the  center  of  a  right  section, 
parallel  to  all  the  elements.  It  is  a  useful  line  in  defining  the 
cylinder  in  space. 

Varieties  of  Cylinders. — The  characteristic  distinction  between 
cylinders  is  the  curve  of  their  right  section.  This  is  at  some 
variance  to  the  cone  distinctions,  which  are  founded  on  base  and 
axial  peculiarities.  The  reason  for  this  is  that  it  is  absolutely 
necessary  to  know  the  right  section  of  a  cylinder  in  most  of  the 
problems  arising;  as,  for  instance,  in  its  development,  the  right 
section  always  rolls  out  in  a  straight  line.  Also,  in  most  tangent 
problems,  the  right  section  has  to  be  considered.  So,  when  a 
cylinder  is  spoken  of  as  circular,  elliptical,  etc.,  a  right  section 
is  understood. 

Limits. — When  the  directrix  becomes  a  straight  line,  the  cylin- 
der becomes  a  plane,  and  when  the  directrix  becomes  a  point  the 
cylinder  becomes  a  line. 

186.  Tangent  Planes  to  Cylinders. — A  plane  which  contains 
two  consecutive  elements  is  tangent.     The  Tangent  Plane  is 
parallel  to  the  axis,  and  to  every  element.     The  two  consecutive 
elements  contained  in  the  tangent  plane  form  the  Element  of 
Contact.     This  element  of  contact  is  one  of  the  two  tangent  lines 
necessary  to  determine  the  tangent  plane.     From  this  fact  we 
derive  the  special  rule  for  passing  planes  tangent  to  cylinders. 

Rule. — A  tangent  line  and  a  line  parallel  to  the  axis,  inter- 
secting the  tangent,  determine  a  tangent  plane  to  a  cylinder. 

Axiom. — When  the  base  of  a  cylinder  is  in  H,  the  .//-trace  of 
any  plane,  tangent  to  the  cylinder,  will  be  tangent  to  the  base 
curve  of  the  cylinder. 

Axiom. — A  plane  cannot  in  general  be  tangent  to  two  cylinders 
unless  the  axes  are  parallel.  The  exceptions,  which  are  not 
frequent,  will  only  be  found  when  two  outside  elements  of  the 
two  cylinders  are  in  the  same  plane.  .That  plane  will  be  the 


SINGLE-CURVED  SURFACES  187 

tangent  plane.  If  two  cylinders  have  parallel  axes,  four  planes 
tangent  to  both  can  usually  be  passed.  Two  cylinders  of  equal 
diameter,  with  axes  intersecting,  can  have  two  common  tangent 
planes  outside. 

187.  Development. — As  was  before  mentioned,  the  right  sec- 
tion of  a  cylinder  rolls  out  into  a  straight  line.     This  is  true,  be- 
cause the  right  section  is  perpendicular  to  all  the  elements,  and 
when  they  are  rolled  out  they  remain  parallel,  and  the  plane 
perpendicular  to  them  cuts  their  plane  in  a  straight  line  per- 
pendicular to  them.     Therefore,  the  first  thing  to  do  is  to  rectify 
a  right  section.     Then  lay  off  perpendiculars,  properly  spaced, 
with  the  true  lengths  of  the  various  elements  laid  off  in  order. 

PROBLEMS  RELATING  TO  CYLINDERS 

188.  Problem  72. — To  locate  a  point  on  the  surface. 
Analysis. — 1.  Assume  one  projection  of  the  point. 

2.  Draw  a  line  through  that  projection  parallel  to  the  axis. 
(This  will  be  the  element  on  which  it  lies.) 

3.  Find  the  intersection  of  this  element  with  the  base. 

4.  Project  this  intersection  to  the  other  projection  of  the  base 
and  draw  the  other  projection  of  the  element. 

5.  Project  the  point  to  the  second  projection  of  the  element. 
Let  the  student  make  the  construction. 

189.  Problem  73. — To  find  the  intersection  of  a  plane  and  a 
cylinder. 

Analysis. — 1.  Draw  a  sufficient  number  of  elements. 

2.  Obtain  the  piercing  points  of  these  elements  with  the  plane. 

3.  Draw  a  smooth  curve  through  these  piercing  points  in  order. 

190.  Problem  74. — To  find  the  points  in  which  a  straight  line 
pierces  a  cylinder. 

There  are  two  methods :  1.  to  use  when  the  base  of  the  cylinder 
lies  in  one  of  the  planes  of  projection;  2.  to  use  when  the  base  is 
in  space. 

Analysis.    First  Method. — When  the  base  is  in  H,  V  or  P. 

1.  Pass  a  plane  through  the  line  parallel  to  the  axis. 

2.  At  the  points  in  which  the  trace  of  this  plane  crosses  the 
base,  draw  elements  of  the  cylinder. 

3.  The  points  in  which  these  elements  are  intersected  by  the 
given  line  will  be  the  required  points. 


188 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Second  Method. — When  the  base  is  in  space. 

1.  Draw  a  sufficient  number  of  elements. 

2.  Pass  a  plane  through  the  line,  preferably  a  projecting  plane. 

3.  Find  the  intersections  of  this  plane  with  enough  of  the  ele- 
ments to  yield  a  curve  that  will  be  intersected  twice  by  the  given 
line. 

4.  The  required  points  will  then  be  where  the  given  line  crosses 
the  curve. 

191.  Problem  75. — To  pass  a  plane  tangent  to  a  cylinder 
through  a  given  point  on  the  surface. 

Analysis. — 1.  Through  the  given  point  draw  a  line  parallel  to 
the  axis. 


(This  will  be  one  line  of  the  required  plane,  the  element  of 
contact.) 

2.  At  the  point  where  the  element  of  contact  crosses  any  curve 
of  the  surface  (preferably  one  of  the  bases),  draw  the  tangent  to 
that  curve. 


SINGLE-CURVED  SURFACES 


189 


3.  Pass  the  plane  through  the  element  and  tangent. 
Construction. — Let  O  (Fig.  141)  be  the  given  point  on  the 
surface  of  the  elliptical  cylinder  whose  axis  is  AB. 

1.  Draw  the  element  OP. 

2.  Draw  the  tangent  to  the  base  at  P. 

3.  This  tangent  will  be  the  //-trace  of  the  tangent  plane. 

4.  Draw  the  F-trace  by  any  of  the  known  means. 

Note. — This  construction,  as  illustrated  in  Fig.  141,  is  only  possible  when 
the  base  of  the  cylinder  is  in  H,  V  or  P.  If  the  base  is  in  an  oblique  plane, 
the  tangent  will  not  be  one  of  the  traces  of  the  plane.  The  analysis  will 
apply  to  any  situation  of  the  cylinder. 

192.  Problem  76. — To  pass  a  plane  tangent  to  a  cylinder 
through  a  point  outside  the  surface. 


FIG.  142. 

Analysis. — 1.  Draw  a  line  through  the  point  parallel  to  the  axis. 

2.  Find  the  point  in  which  this  line  pierces  the  plane  of  the 
base. 

3.  From  this  piercing  point  draw  a  tangent  to  the  base. 

4.  The  parallel  line  and  the  tangent  line  will  determine  the 
required  tangent  plane. 

Let  the  student  make  his  own  construction  for  a  cylinder  similar 
to  Fig.  141.     The  construction  here  given  is  for  a  difficult  case. 

Note. — The  projections  of  the  cylinder  are  not  drawn.     This  is  done  so  as 
not  to  complicate  the  figure  unnecessarily. 

Construction. — Let  the  cylinder  be  a  circular  cylinder  of  %-in. 


190 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


diameter,  whose  axis  is  AB  (Fig.  142).    Let  it  be  required  to 
pass  a  tangent  plane  through  M,  a  point  outside  the  surface. 

1.  Draw  the  line  MN,  parallel  to  AB  through  M. 

2.  Pass  a  plane  Q  through  A,  perpendicular  to  AB.     (This 
will  be  the  plane  of  the  base,  cutting  a  circle  from  the  surface.) 

3.  Find  the  piercing  point,  N,  of  MN  with  Q. 

4.  Revolve  A  and  N  into  H  about  HQ. 

5.  Draw  a  %-in.  circle  about  ai  as  a  center. 

6.  From  ni  draw  a  tangent  to  the  circle. 

7.  Counter-revolve  the  tangent  nio  into  the  plane  Q. 

8.  Pass  the  plane  S  through  NO  and  MN.     It  will  be  the 
required  tangent  plane. 


FIG.  143. 

193.  Problem  77. — To  pass  a  plane  tangent  to  a  cylinder  and 
parallel  to  a  given  line. 

Analysis. — 1.  Pass  a  plane  parallel  to  the  axis  and  the  given 
line.  Why? 

2.  Find  the  intersection  of  this  plane  and  the  plane  of  the  base. 

3.  Draw  a  tangent  to  the  base  parallel  to  the  intersection 
obtained  in  operation  (2). 

4.  Pass  the  required  tangent  plane  through  this  tangent  line, 
parallel  to  the  auxiliary  plane,  passed  in  operation  (1). 

Construction. — Let  AB,  Fig.  143,  be  the  axis  of  the  cylinder, 
and  let  it  be  required  to  pass  the  tangent  plane  parallel  to  XY. 


SINGLE-CURVED   SURFACES 


191 


1.  Pass  the  auxiliary  plane  Q  through  AB  parallel  to  XY. 

2.  Draw  the  required  plane  T,  tangent  to  the  cylinder  and 
parallel  to  Q.     The  //"-trace  of  T  will  be  tangent  to  the  base  in 
H,  and  parallel  to  HQ,  and  the  F-trace  of  T  will  be  parallel  to  VQ. 

194.  If  the  cylinder  be  circular  and  its  axis  oblique,  the  best 
method  for  the  foregoing  problem  is  the  "Normal"  method. 
Its  analysis  is  as  follows: 

1.  Pass  a  plane  through  the  axis  parallel  to  the  given  line. 

2.  Pass  a  plane  parallel  to  this  plane  at  a  distance  equal  to  the 
radius  of  the  cylinder. 


FIG.  144. 

The  problem  may  be  done  in  this  way  without  drawing  any 
elements  or  bases  of  the  cylinder.  Let  the  student  prove  that 
the  "normal"  method  is  correct. 

195.  Problem  78. — To  pass  a  plane,  inclined  at  any  given  angle 
to  H  or  V,  tangent  to  a  cylinder. 

Limitations. — The  limits  of  this  angle  must  be  between  90° 
and  the  angle  between  the  elements  and  the  stipulated  plane. 

Analysis. — 1.  Pass  an  auxiliary  plane  through  the  axis  making 
the  given  angle  with  H  or  V,  as  required;  see  Problem  40. 


192 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  If  the  base  of  the  cylinder  is  in  H,  draw  the  //-trace  of  the 
required  plane,  tangent  to  the  base  and  parallel  to  the  //-trace 
of  the  auxiliary  plane. 

3.  Draw  the  F-trace  of  the  required  plane  parallel  to  the  V- 
trace  of  the  auxiliary  plane. 

196.  Normal  Method. — When  this  problem  is  required  of  a 
circular  cylinder,  oblique  to  H  and  V,  the  normal  method,  like 
that  in  Art.  194,  is  the  easiest  and  best. 

Fig.  144  shows  the  construction  of  such  a  problem,  in  which 
AB  is  the  axis  and  AC  the  radius  of  the  given  cylinder.  The 
elements  of  the  auxiliary  cone  make  the  given  angle  with  H. 
T  is  one  of  the  four  possible  required  planes. 


o 

Develooment 
FIQ.  145. 

197.  When  the  axis  is  parallel  to  H,  and  the  angle  between 
the  tangent  plane  and  H  is  given,  the  problem  is  very  easily 
solved  by  projecting  on  a  plane  perpendicular  to  the  axis. 

Note. — Four  planes  can  be  passed  tangent  to  any  cylinder,  making  the 
given  angle,  if  the  problem  is  possible. 

198.  Problem  79.— To  develop  the  surface  of  a  cylinder. 
Construction. — To  develop  the  elliptical  cylinder,  whose  axis 

is  AB  (Fig.  145). 

1.  Draw  a  sufficient  number  of  elements. 

2.  Pass  a  plane  perpendicular  to  the  axis,  as,  for  example, 
along  the  line  CD. 

3.  Revolve  this  right  section  to  show  its  true  size. 

4.  Lay  off  the  base  line,  CiDj,  equal  in  length  to  the  length  of 


SINGLE-CURVED   SURFACES 


193 


the  ellipse,  with  divisions  equal  to  the  respective  arcs,  as  shown 
by  the  corresponding  numerals. 

5.  Through  all  these  divisions  draw  perpendiculars  to  CiDi. 

6.  Lay  off  on  these  perpendiculars  the  true  lengths  of  their 
respective    elements,    measured    above    and    below    the    right 
section. 

7.  Draw  a  smooth  curve  through  the  extremities. 

Note. — It  is  not  always  necessary  to  show  both  projections  of  the  right 
section.     In  this  case  the  //-projection  is  unnecessary. 


PIPE  ELBOWS 

199.  The  object  of  a  pipe  elbow  is  to  make  a  turn  through  any 
desired  angle  and  preserve  the  pipe  diameter  through  the  turn. 


FIG.  146. — Five-piece  pipe  elbow. 

The  surface  thus  approximated  is  that  of  the  torus,  and,  in  case 
of  a  double  bend,  the  serpentine. 

The  elbow  is  laid  out  as  follows: 

Divide  the  center-angle  (0)  by  2n  —  2  (n  being  the  number  of 
pieces  in  the  turn).  This  division  will  yield  a,  the  angle  of  the 

joint,  i.e.,  a  ~ — —^ 

Example. — Required  to  design  a  5-piece  elbow  for  a  90°  turn 
(axial  angle),  on  10-in.  radius. 

The  center-angle  0  is  the  supplement  of  the  axial  angle;  .'.  0  = 
90°. 

2n  -  2  =  8.     .'.  a  =  \\Y±. 

18 


194 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


(See  Fig.  146.)  Lay  off  a  90°  arc  on  a  10-in.  radius,  and  divide 
the  arc  into  arcs  of  11^°.  The  first  of  these  radii  will  be  the 
projection  of  the  first  joint,  and  the  odd  radii,  the  third,  fifth, 
and  seventh,  will  be  the  projections  of  the  other  joints.  Pipes 
laid  out  perpendicularly  to  the  even  radii  will  give  a  pipe  bend 
of  uniform  right  section. 


200. 


EXERCISES 


V     L.V 
dtP 


iah 


Graphic  Layout  No.  55.  —  Axes  for  circular  or  elliptical  cylinders,  with 
bases  perpendicular  to  the  axes. 

Bases.  —  Circles  1  in.,  IK  in.,  2  in.;  ellipses  1  in.  X  IK  in.,  1  in.  X  2  in. 
IK  in.  X  2  in.,  l^.,n.  X  2K  in. 

Altitudes.—  IK  in.,  2  in.,  2K  in.,  3  in. 

Note.  —  The  signs  for  these  lines  may  be  reversed,  and  the  cylinders  drawn 
in  III,  if  desired. 

Dimensioned  Layout.  —  Center  lines  for  circular  cylinders  or  elliptical 
cylinders,  with  bases  perpendicular  to  the  axes. 


(1)  A(3,  0  -  IK)  B(3  +  3  -  IK). 

(2)  A(3  +  1,  0)  B(3  +  1  -  3). 

(3)  A(2  +  IK  -  1)  B(4  +  IK  ~  1). 

(4)  A(2  +  1  -  1)  B(4  +  1  -  2). 

(5)  A(2  +  2  -  1)  B(4  +  1  -  1). 

(6)  A(3  +  2  -  K)  B(3  +  1  -  2). 

(7)  A(3  +  2K  -  2)  B(3  +  1  -  1). 

(8)  A(l  +  2K  -  1)  B(3K  +  1-2). 

(9)  A  (2  +  2  -  2K)  B(4  +  1  -  1). 

Note.  —  The  foregoing  axes  may  be  drawn  in  777  by  reversing  the  signs  + 
and  —  .  In  this  event,  the  signs  on  the  plane  traces  and  piercing  lines 
should  be  reversed,  if  they  enter  the  exercise. 


SINGLE-CURVED  SURFACES 


195 


1 

c* 

2\^ 

J 

Y 

4 
VM 

V 

^<" 

x^ 

HM; 

5 

^^ 
••• 

HN 
~^7fs]" 

6 

X 

7          V 

8  f 

X 

\ 

Graphic  Layout  No.  66.  —  Cutting  planes.     (Note.  —  If  the  cylinders  are 
drawn  in  ///,  reverse  the  lettering  of  these  traces.) 

1     c* 

p* 

^ 

V 

C 
dV 

3             1^             4    v                v      5  jC 

c            d          1^^^     v 

!d"            1    !  _] 

A 

JdH 

J 

id"       r  x" 

lcH           c"         d"            ^Xid" 

Graphic  Layout  No.  57.  —  Piercing  lines.  (Note.  —  If  the  axes  are  drawn 
in  III,  draw  these  lines  in  ///,  and  in  any  case,  draw  them  so  that  they  are 
likely  to  pierce  the  cylinder. 

Cutting  Planes. 

(1)  QUM  +  3)  1K(5  -  2). 

(2)  R(l  +  2)  4(4  -  3). 

(3)  S(5  +  3)  1(5  -  2). 

(4)  T(l  '+  3)  2(5  -  2). 

(5)  U(+2)«(-l). 

(6)  W(  +  M)«(+3): 

(7)  X(l  +  1)  5(1  +  2^). 

(8)  Y(l  +  3)  3(5  -  3). 

(9)  Z(4  +  3)  3(4  -  2). 

(10)  P(3K+3)3^(3K  -3). 


Piercing  Lines. 

(1)  A(2,  0  -  1M)  B(4  +  2  -  2). 

(2)  C(2,  0  -  2)  D(4  +  2  -  ^). 

(3)  E(3  +  2  -  K)  F(3  +  1  - 

(4)  G(3  +  2  -  2)  K(3  +  K  -  1) 

(5)  M(l  +  1  -  1M)  N(4  + 


792.  Cylinder ,  axis ,  base,  or  right  section ;  locate  the  projec- 


tions  of  a  point  on  the 


(front  or  rear)  surface. 


196 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


793.  Cylinder ,  axis ,  base,  or  right  section ;  draw  the  inter- 


section, and  its  true  size,  made  by  plane 


794.  Draw  the  development  of  the  cylinder  in  Ex.  793,  included  between  one 
of  the  bases  and  the  cutting  plane. 

795.  Cylinder ,  axis ,  base,  or  right  section ;  find  the  points 


in  which  it  is  pierced  by  line 


796.  Draw  the  shortest  line  on  the  surface  between  the  two  piercing  points 
found  in  Ex.  795. 

797.  Draw  the  H-,  V-,  and  P-intersections  of  cylinder ,  axis ,  base, 

or  right  section ,  continuing  the  cylinder  to  meet  the  planes,  if 

necessary. 

798.  Draw  the  traces  of  the  plane  tangent  to  cylinder ,  axis ,  base, 

or  right  section ,  at  a  point  assumed  on  the (front  or  rear) 

surface, in.  from  H. 

799.  Draw  the  traces  of (one  or  two)  tangent  planes  to  cylinder , 

axis ,  base,  or  right  section ,  through   a   point  without  the 

surface,  in (/,  //,  ///  or  IV). 

800.  Draw  the  traces  of (one  or  two)  tangent  planes  to  cylinder , 

axis    ,    base,      or    right    section ,    parallel    to    A(  ) 

B(  ). 


801.  Draw  the  traces  of 
cylinder ,    axis  - 


—   (one,  two,  three,  four)  tangent  planes  to 
-,    base,    or   right   section,    inclined  °  to 


(H,  V,  or  P). 


Base  in  H 


a 

Base  inH 


Base  in  V 


iar 


Base  in  V 


bv 

BaseinV 


Base  inPafA 


u"l^^ 

°  "Base  in  Pat  A 


Graphic  Layout  No.  68. — Axes  for  elliptical  cylinders,  with  circular  or 
elliptical  bases  in  H,  V,  or  P,  as  given. 

Bases. — 1  in.,  1%  in.,  1%  in.,  2  in.,  circles;  1  in.  X  IK  in.,  1  in.  X 
2  in.,  IK  in-  X  2  in.,  ellipses. 

Dimensioned  Layout  for  Axes  of  Foregoing  Cylinders. 

(1)  A(1K,  0  -  IK)  B(3K  +  2  -  IK).     Base  in  H. 

(2)  A(2  +  IK,  0)  B(4  +  K  -  2).     Base  in  V. 

(3)  A(3  +  2,  0)  B(3,  0  -  IK)-     Base  in  H. 

(4)  A(3  +  1%,  0)  B(3,  0  -  2).     Base  in  V. 


SINGLE-CURVED  SURFACES 


197 


(5)  A(l  +  2  -  1)  B(3>£  +  1-1).     Base  in  P,  at  B. 

(6)  A(1K,  0  -  1)  B(3  +  2  -  2).     Base  in  H. 

(7)  A(l  +  2M  -  2)  B(3H  +  1-1).     Base  in  P,  at  B. 

Note. — The  axes  in  both  layouts  may  be  located  in  III  by  reversing  the 
+  and  —  signs,  at  the  option  of  the  instructor. 

Note  2. — The  cutting  planes  and  piercing  lines  for  the  following  exercises 
are  found  in  Graphic  Layouts  Nos.  56  and  57. 


802.  Cylinder,  axis 
rear)  surface  — 

803.  Cylinder,  axis 


base 


assume  a  point  on  the 


(front  or 


in.  from  H. 
— ,  base ;  draw  the  projections  and  true  size  of 


the  section  made  by  plane 
804.  Cylinder,  axis ,  base  - 


-;  draw  the  projections  and  true  size  of 
its  right  section  at  the  middle  point  of  the  axis. 

805.  Cylinder,  axis •,  base   ;    locate    the    points    in   which  it  is 

pierced  by  line . 

806.  Draw  the  shortest  line  on  the  surface  of  the  cylinder  between  the 
points  found  in  Ex.  805. 

807.  Draw  the  H-,  V-,  and  P-sections   of  cylinder,  base ,   axis . 

(Continue  the  surface  until  the  section  is  made.) 

808.  Cylinder,  axis ,  base ;  draw  the  traces  of  the  plane  tangent 


at  a  point  on  the 
809.  Cylinder,  axis  — 


-  (front  or  rear)  surface,  assumed, 
base :  draw  the  traces  of  — 


(one,  two) 

planes  tangent  through  a  point  outside  the  surface,  assumed  in  — 
(/,  II,  III,  IV). 

810.  Cylinder,  axis ,  base ;  draw  the  traces  of (one,  two,  three, 

four)  planes  tangent,  inclined °  to (H,  V,  or  P). 

811.  Cylinder,  axis -,  base ;  draw  the  traces  of (one,  two)  planes 

tangent  to  it  and  parallel  to  X(  )  Y(  ). 

PRACTICAL  EXERCISES 
I* 


k-A->U-C->J 


\ 


A 

B 

(, 

D 

2 

83EBE 

z 

3 

4 

4 

5 

6 

6 

7 

Q 

Q 

$ 

id 

10 

n 

k—  A—-- » 


CO 


A 

B 

C 

P 

t 

f 

/    22 

10 

12 

<? 

3 

12 

2    24 

1? 

t't 

6 

$ 

12 

3    26 

14 

14 

7 

10 

12 

4     28 

16 

14 

7 

10 

12 

2-Piece  Pipe  Elbow   Helical  Tubing         Locomotive  Stack 

Graphic  Layout  No.  59. — Dimensions  in  inches. 


812.  Lay  out  the  patterns  for  Two-piece  Pipe  Elbow 

813.  Lay  out  the  patterns  for  Helical  Tubing . 

814.  Lay  out  the  patterns  for  Locomotive  Stack 


0  = 


198 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


H 


Oblong 


Number  of  Pieces 

H 

R 

w 

H 

/ 

3,4,5,6,7 

60°75°,90?/OS?/ZO° 

8 

4 

•t 

77 

10 

5 

2 

6 

3 

n 

12 

6 

2 

8 

I 

» 

16 

8 

3 

I 

5 

7 

20 

10 

S 

ft 

6 

> 

7 

4 

2 

5 

/ 

7 

9 

5 

2 

7 

8 

7 

II 

6 

^ 

5 

y 

> 

•> 

14 

8 

3 

5 

10 

> 

> 

18 

IO 

4 

8 

Multi -Joint  Pipe  Elbow 

Graphic  Layout  No.  60. — Dimensions  in  inches. 


815.  Lay  out  the  patterns  for  the  Multi-joint  Pipe  Elbow 
number  of  pieces . 

816.  Lay  out  the  patterns  for  the  Oblong  Pipe  Elbow 
number  of  pieces . 


0  = 


Corrugated    Pipe. 
Graphic  Layout  No.  61. 


Scoop  for  Scale. 


817.  Lay  out  the  patterns  for  the  Corrugated  Pipe. 

818.  Lay  out  the  patterns  for  the  Scoop. 


SINGLE-CURVED  SURFACES 


199 


A 

B 

8 

/ 

1 

7 

45'6D:67J£,75° 

•I 

4 

7 

n 

I 

4 

B 

a 

4 

6 

8 

5 

6 

6 

IO 

» 

(5 

6 

12 

1) 

7 

8 

I! 

1> 

fi 

$ 

14 

» 

9 

8 

16 

» 

1*0*1 


Oblique  Connector 
for  Circular  Pipe 


A 

B 

C 

D 

E 

F 

G 

K 

89# 

109 

80 

8 

64 

2O 

SO 

36 

7J^> 

89'/s 

64 

0 

53h 

16 

43 

32 

$//£ 

74 

S3 

7 

44 

13 

38 

?8 

41 

49 

35 

S 

?9 

5 

f4 

S4 

^l 

?3 

IT/i 

3 

Mi 

3'/4 

17 

16 

Morse  Rarefied  Dust  Collector 


Graphic  Layout  No.  62. — Dimensions  in  inches. 

819.  Lay  out  the  patterns  for  Oblique  Connector  - 

820.  Lay  out  the  patterns  for  the  Dust  Collector  - 


K-B--H 


Offset  Pipe  Reducer 
Graphic  Layout  No.  63. — Dimensions  in  inches. 


A 

B 

C 

16 

10 

IO 

t8 

10 

10 

20 

11 

/? 

24 

12 

tS 

Z6 

/2 

14 

?8 

14 

14 

30 

14 

16 

32 

14 

16 

36 

16 

18 

40 

16 

18 

Oblong  Hood 


820.  Lay  out  the  patterns  for  Offset  Pipe  Reducer 

821.  Lay  out  the  patterns  for  Oblong  Hood  . 


200 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


THE  CONVOLUTE 


201.  The  third  class  of  single-curved  surfaces,  the  Convolute, 
is  generated  by  the  tangent  line  to  any  space  curve,  as  it  moves 
along  the  curve.     The  tangent  line  to  a  curve  contains  two  con- 
secutive points  of  the  curve.     This  means  that  two  consecutive 
tangents  have  one  point  in  common;  thus,  A,  B,  and  C  are  con- 
secutive points  in  a  curve,  and  AB  and  BC  are  therefore  two 
consecutive  tangents  with  B   as  their  intersection.     Thus  all 
consecutive  tangents  to  any  space  curve  intersect  two  and  two, 
and  therefore  become  elements  of  a  single-curved  surface. 

202.  Helical  Convolute. — The  only  convolute  of  importance  is 
that  generated  by  the  tangent  to  the  cylindrical  helix.     Fig.  147 
shows  both  projections. 


FIG.  147. — Projections  of  a  convolute. 

It  is  sometimes  called  the  "Developable  Helicoid,"  and  its 
most  frequent  application  is  in  screw  conveyors. 

Theoretically,  the  surface  is  unlimited,  the  portion  shown  here 
being  a  "three-quarter  turn."  In  practical  design  it  is  always 
limited  by  a  cylinder  and  upper  and  lower  bases,  or  by  two  cylin- 
ders. The  //-trace  of  the  surface  is  an  involute.  (See  Fig.  147.) 

203.  Problems. — The  problems  that  relate  to  this  surface  do 
not  differ  materially  from  those  of  other  single-curved  surfaces; 
therefore  the  only  problem  given  a  solution  here  is  the  problem 
of  the  development  of  the  surface. 


SINGLE-CURVED  SURFACES 


201 


204.  Problem  80.— To  develop  the  surface  of  a  convolute. 

Discussion. — If  the  helical  directrix  be  rolled  into  a  plane, 
element  by  element,  it  will  be  seen  to  have  a  constant  radius  of 
curvature,  and  will  therefore  develop  as  a  circle.  This  radius 
is  the  normal  line  at  any  point  of  the  helix  that  intersects  the 
axis,  because  all  such  normals  will  have  equal  lengths.  The 
elements  of  the  convolute,  being  tangent  to  the  helix,  are  per- 
pendicular to  the  respective  normals,  and  when  the  rolling  out 
occurs,  these  elements  will  be  consequently  tangent  to  the  circle 
made  by  developing  the  helix.  As  the  lengths  of  the  tangents 
are  proportional  to  their  //"-projections,  the  developed  curve  of 
their  extremities  will  be  the  involute  to  the  developed  helix. 


The  Convolute  and  its  Development. 
FIG.  148. 


This  is  so  because  each  tangent  is  equal  in  length  to  the  helix 
from  the  origin  to  the  point  of  tangency,  and  when  the  helix  is 
developed  into  a  circle,  the  tangents  are  equal  to  the  arc  from  the 
origin  to  the  point  of  tangency.  This  corresponds  to  the  genera- 
tion of  the  involute. 

Analysis. — The  entire  analysis  is  practically  confined  to  the 
determination  of  the  radius  of  the  circle  of  the  developed  helix. 

Graphical  Solution. — To  determine  the  development  radius 
graphically,  erect  a  perpendicular  cvdv  (Fig.  148)  to  the  tangent 
bvfv  at  the  point  where  bvfv  crosses  the  outside  element  of  the 
cylinder.  The  horizontal  line  bvcv  included  between  this  per- 
pendicular and  the  point  of  tangency  is  the  desired  radius. 


202  PRACTICAL  DESCRIPTIVE  GEOMETRY 

Construction  of  the  Development.  —  Draw  the  circle  with  the 
radius  bvcv  and  draw  an  involute  to  it.  Make  the  arc  of  the 
developed  involute  equal  to  the  arc  of  the  helix,  since  the  helix 
from  any  point  to  the  origin  is  equal  to  the  tangent  from  that 
point  to  the  plane  of  the  origin.  This,  then,  is  accomplished  by 
measuring  the  tangent  and  laying  off  its  length  on  the  involute 
in  the  development. 

Mathematical  Value.  —  The  mathematical  value  for  the  radius 

•p 

of  the  development  is  the  expression  —  ^>  in  which  R  is  the 

radius  of  the  helical  cylinder,  and  6  is  the  inclination  of  the  ele- 
ments to  the  horizontal. 

EXERCISES 

822.  Draw  a  270°  convolute  to  the  helix,  whose  axis  is  A(l>£,  0  —  1J4) 
B(l^  +  2  —  134),  1-in.  diameter,  2-in.  pitch.     Find  the  traces  of  the 
tangent  plane  containing  C(2^,  x  —  2)  on  the  surface. 

823.  Draw  the  convolute  in  Ex.  822.     Draw  the  traces  of  the  tangent  plane 
through  D(l%  +  Y±,  y)  on  the  front  surface. 

824.  Draw  a  270°  convolute  to  the  "left-hand"  helix,  whose  axis  is  E(4^, 
0-1)  F(4^  +3  -  1),   1^-in.  diameter,  2%-ia.  pitch.     Draw  the 
traces  of  a  plane  tangent  at  the  point  G(4%,  x  —  1%)  on  the  surface. 

825.  Draw  the  convolute  in  Ex.  824.     Draw  the  traces  of  the  plane  tangent 
at  K(3^  +  1,  y)  on  the  front  surface. 

826.  Draw  the  convolute  in  Ex.  822.     Draw  the  traces  of  a  plane  tangent 
containing  M(l  +  2  —  2^). 

827.  Convolute  in  Ex.  824.     Draw  the  traces  of  the  tangent  plane  con- 
taining N(4K  +  1  -  234). 

828.  Convolute  in  Ex.  822.     Find  its  intersection  with  the  plane  S(l  +  1) 
4(1  -  3). 

829.  Convolute  in  Ex.  824.     Find  its  intersection  with  the  plane  T(+ 


830.  Develop  the  surface  of  the  convolute  in  Ex.  822. 

831.  Develop  the  surface  of  the  convolute  in  Ex.   824. 

832.  Find  the  shortest  path  on  the  surface  of  the  convolute  in  Ex.  822  from 
0(1%,  x  -  2)  to  P(3,  y  -  1). 

833.  Find  the  shortest  path  on  the  surface  of  the  convolute  in  Ex.  824  from 
C(5,  x  -  1,34)  to  D(3,  y  -  M). 


CHAPTER  VII 
WARPED  SURFACES 

204.  Discussion. — According  to  the  definition  in  Art.  167,  a 
warped  surface  is  generated  by  moving  a  straight  line  so  that  no 
two  consecutive  positions  shall  be  in  the  same  plane.  It  thus 
falls  into  the  classification  with  single-curved  surfaces  as  a  ruled 
surface.  Planes  may  be  passed  through  any  point  of  a  warped 
surface  that  will  cut  straight  lines  from  the  surface,  always  one, 
and,  in  the  case  of  several  surfaces,  two  straight  lines.  Other 
planes  cut  curves  from  them. 

Warped  surfaces  are  of  many  varieties  and  are  common  in 
practical  work.  In  many  cases,  when  they  occur  in  practical 
work,  it  is  not  necessary  for  the  draftsman  to  recognize  their 
peculiarities,  but  very  often,  as  in  arches,  sheet  metal  work, 
turbine  blades,  propeller  blades,  skew  gears,  conveyors,  etc.,  it  is 
highly  important  that  engineers  and  draftsmen  know  their  prop- 
erties and  methods  of  generation. 

206.  Methods  of  Generation. — Warped  surfaces  may  be 
generated: 

1.  By  revolving  one  line  about  another  not  in  the  same  plane. 

2.  By  moving  a  straight  line,  the  generatrix,  along  two  lines, 
directrices,  either  straight,  curved,  or  both,  keeping  parallel  to 
some  plane,  called  the  Plane  Director. 

3.  By  moving  a  straight  line  along  three  directrices,  straight, 
curved,  or  both. 

4.  By  moving  a  straight  line  along  an  axis  at  a  constant  angle 
with  the  axis,  following  a  space  curve  as  a  directrix. 

Other  methods  might  be  invented,  but  these  cover  all  the  cases 
met  in  practical  work. 

206.  Tangent  Planes. — In  Art.  170  we  find  the  general  rule 
for  tangent  planes  to  all  Ruled  Surfaces,  as  follows :  One  tangent 
line  and  the  element  at  any  point  determine  the  tangent  plane 
at  that  point. 

In  general,  a  tangent  plane  is  only  tangent  at  one  point  of  a 
warped  surface.  There  are  certain  points,  as  in  conoids  and 
warped  cones,  where  the  tangent  plane  will  be  tangent  all  along 
the  element,  but  such  exceptions  will  be  readily  recognized. 

203 


204  PRACTICAL  DESCRIPTIVE  GEOMETRY 

All  planes  tangent  to  warped  surfaces  (except  in  the  cases  just 
noted)  intersect  the  surface,  because  they  are  not  tangent  all 
along  the  element,  and,  conversely,  any  plane  that  intersects  a 
warped  surface  in  an  element  will,  in  general,  be  tangent  to  the 
surface  at  some  point. 

207.  Double-Ruled  Surfaces. — Two  surfaces,  the  Hyperboloid 
of  Revolution  of  One  Nappe  and  the  Hyperbolic  Paraboloid,  are 
called  double-ruled  surfaces  from  the  fact,  as  will  be  shown  later, 
that  they  are  capable  of  generation  by  two  different  straight 
lines.     This  yields  the  phenomenon  that  through  every  point 
in  the  surface  two  straight  lines  (elements)  can  be  drawn  lying 
wholly  within  the  surface.     From  this  fact  we  have  the  special 
rule  for  tangent  planes  to  these  surfaces:  The  two  elements  of  a 
double-ruled  surface  through  any  point  of  the  surface  determine 
the  tangent  plane  at  that  point. 

208.  Representation. — Some  warped  surfaces  can  be  shown 
in  outline  in  orthographic  projection  with  sufficient  clarity,  and 
others  must  be  shown  by  drawing  a  sufficient  number  of  elements. 
In  practical  drafting,  where  the  surface  is  always  limited,  the  out- 
lines are  usually  sufficient. 

209.  Location  of  Points. — To  locate  a  point  on  any  warped 
surface,  one  projection  being  given,  find  the  piercing  point  of 
the  projector  of  the  point  with  the  surface.     This  is  done  by 
passing  a  plane  through  the  projector,  and  finding  its  inter- 
section with  the  surface.     Where  the  projector  intersects  this 
line  will  be  the  location  of  the  point.     In  many  cases  it  will  be 
easy  to  draw  the  element  through  the  point,  simplifying  the 
operation. 

210.  Developments. — Strictly    speaking,    a    warped    surface 
cannot  be  developed,  because  no  two  consecutive  elements  lie  in 
the  same  plane,  and  cannot,  therefore,  be  rolled  into  a  plane 
without  "warping"  them;  that  is,  destroying  the  relation  be- 
tween them.     However,   there  are  approximate   developments 
that  give  very  satisfactory  results,  almost  as  good  as  the  devel- 
opment of  a  cone  by  triangulation.     These  warped  developments 
also   are   obtained   by   triangulation.     A   sufficient  number   of 
elements  are  drawn,  dividing  the  surface  into  minute  warped 
quadrilaterals,  in  which  the  amount  of  warp  is  so  slight  as  to  be 
negligible.     Diagonals   are   then   drawn   across   each   of   these 
quadrilaterals,    dividing  them  into  triangles.     These  triangles 
are  then  measured  and  placed  in  order  in  their  (approximate) 


WARPED  SURFACES 


205 


true  size.  The  resulting  surface,  obtained  by  bending  back 
this  development,  cannot  be  the  exact  surface,  as  shown  in  the 
drawing,  but  will  be  near  enough  for  practical  purposes.  In 
making  single-curved  surfaces  from  flat  patterns  there  need  be 
no  distortion  of  the  metal,  but  in  making  up  a  warped  surface, 
there  will  be  some  distortion. 

THE  HELICOID 

211.  The  most  frequent  warped  surface  encountered  in  draft- 
ing is  the  Helicoid.  It  is  generated  by  Method  4,  Art.  205, 
by  moving  a  straight  line  along  a 
helix  and  its  axis,  keeping  a  constant 
angle  with  the  axis.  The  helix  and 
its  axis  are  the  directrices.  The  sur- 
face is  theoretically  unlimited,  but  in 
practice  is  usually  included  between 
two  helices.  Any  circular  cylinder, 
having  its  axis  coincident  with  that 
of  the  helical  directrix,  will  intersect 
the  helicoid  in  a  helix.  Any  plane 
perpendicular  to  the  axis  will  cut  the 
helicoid  in  a  spiral  of  Archimedes. 
Any  plane  passed  through  the  axis 
will  intersect  the  helicoid  in  an  ele- 
ment. 

Fig.  149  shows  a  helicoid,  with  its 
axial  angle  30°.  In  drawing  the  helix 
a  certain  number  of  points  were  laid 
out,  and  through  these  points  lines 

intersecting  the  axis  are  drawn,  making  the  given  angle  with 
the  axis.  As  the  helix  is  everywhere  equally  distant  from  the 
axis,  this  is  simple,  as  follows. 

1.  Draw  one  element  parallel  to  V,  which  will  be  drawn  in  its 
true  angle  with  the  axis. 

2.  As  the  generatrix  is  moved,  it  climbs  both  axis  and  helix 
in  exactly  the  same  upward  distances.     Therefore,  from  any 
point  on  the  helix,  an  element  may  be  drawn  to  a  point  on  the 
axis  that  is  the  same  distance  up,  measured  on  the  F-projection. 
Thus  in  Fig.  149,  the  point  bv  is  the  same  distance  above  av  as 
the  point  2  is  above  1  on  the  axis. 

The  point  P  is  located  on  the  surface  by  drawing  the  element 


Helicoid. 
FIG.  149. 


206 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


through  the  point  and  locating  the  other  projection  on  the 
element.  If  the  7-projection  of  the  point  is  given,  pass  a  plane 
through  the  point  parallel  to  H.  This  will  cut  the  elements  in  a 
curve  on  which  the  //-projection  will  be  located. 

The  helicoid  is  the  surface  of  the  V-thread,  the  square  thread, 

spiral  springs,  screw  conveyors, 
certain  fire-escapes,  the  twist  drill, 
the  spiral  runway,  helicoidal  arch, 
and  numerous  other  surfaces  fre- 
quent in  machine  and  structural 
design. 

Helicoidal  Arch. — In  the  design 
of  skew  arches,  that  is,  arches  with 
their  axes  oblique  to  one  or  both 
of  the  openings,  warped  surfaces 
are  often  encountered,  cylindroids 
or  cow's  horns  being  the  usual  sur- 
faces in  such  arches.  The  arch 
called  the  helicoidal  arch  takes  its 
name  not  from  the  character  of  its 
surface,  as  might  be  supposed,  but 
from  the  surface  of  the  "coursing 
joints."  The  line  of  the  joint  is 
laid  out  as  a  straight  line  on  the 
developed  surface  of  the  arch,  and, 
when  brought  into  its  actual  posi- 
tion, it  becomes  a  helix.  The  sur- 
face of  the  joint  is  generated  by  a 
line  normal  to  the  surface  of  the 
arch  moving  along  the  helix,  radial  to  the  axis  of  the  helix.  This 
surface  is  therefore  identified  as  a  helicoid.  Inasmuch  as  the 
developed  helix  is  a  straight  line  -the  work  of  laying  out  the 
blocks  is  very  simple.  The  seam,  being  helicoidal,  must  be  laid 
out  with  care,  to  ensure  good-fitting  blocks. 

THE  HYPERBOLOID  OF  REVOLUTION  OF  ONE  NAPPE 

212.  This  peculiar  surface  is  the  only  warped  surface  that  is 
a  surf  ace  of  revolution.  It  is  generated  by  Method  1,  Art.  205, 
by  revolving  a  straight  line  about  an  axis  not  in  the  same  plane. 
Fig.  150  shows. this  surface  represented  by  its  elements.  Either 
projection  of  all  the  elements  could  be  inverted,  and  there  would 


Hyperboloid  of  Revolution. 
Fid.  150. 


WARPED  SURFACES 


207 


be  no  change  in  the  surface.  If  AB  be  considered  an  element,  its 
F-projection  might  be  either  of  the  lines  marked  avbv,  and 
the  same  surface  would  be  generated  by  revolving  either  line 
about  the  axis.  This  makes  it  a  double-ruled  surface,  and  a 
tangent  plane  is  determined  at  any  point  by  two  elements  through 
that  point. 

213.  To  Draw  the  Elements. — All  points  in  the  generatrix  move 
in  circles,  hence  there  is  a  vertex  to  the  curve  of  the  outline,  a 
smallest  circle,  called  the  Gorge  Circle,  generated  by  the  point 
nearest  the  axis.  All  elements  intersect  the  gorge  circle  and  their 
H -projections  are  all  tangent  to  its  H-projection.  Two  limiting 
planes,  bases,  are  always  necessary,  always  taken  perpendicular 


FIQ.  151. — Rolling  hyperboloids.  (Both  surfaces  are  generated  by  re- 
volving the  same  straight  line  about  two  different  axes,  and  they  thus  have 
one  element  in  common,  and  are  tangent  to  each  other  along  this  element.) 

to  the  axis,  and  usually  at  equal  distances  from  the  gorge  circle. 
Being  perpendicular  to  the  axis,  these  bases  are  circular.  Why? 
One  base  is  usually  in  H,  and  the  elements  terminate  in  the  bases. 
Hence,  1.  Draw  the  gorge  and  base  circles.  2.  Draw  as  many 
elements  as  desirable,  16  or  24,  the  //-projections  tangent  to  the 
gorge  circle  and  terminating  in  the  base  circles.  3.  Project  the 
terminals  of  the  elements  to  the  F-traces  of  the  respective  base 
planes. 

Note. — When  the  bases  are  symmetrically  placed,  there  will  be  only  one 
circle  projected  on  #,  as  in  Fig.  150, 

214.  To  Pass  a  Plane  Tangent  at  any  Point  on  the  Surface. — 
Draw  the  two  elements  through  the  given  point.  They  will  de- 
termine the  tangent  plane. 


208 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Tangent  planes  to  this  surface,  as  to  all  surfaces  of  revolution, 
are  perpendicular  to  the  plane  of  the  axis  and  the  point  of  tan- 
gency,  called  the  Meridian  Plane. 

The  Meridian  Section  of  this  surface  is  a  hyperbola,  and  the 
same  surface  can  be  generated  by  revolving  a  hyperbola  about 
its  vertical  axis.  If  proof  of  this  is  desired,  it  can  be  found  in 
Church  and  Bartlett,  page  150. 

The  most  important  function  of  the  hyperboloid  is  found 
in  the  pitch  surfaces  of  skew  gears.  If  one  line  be  successively 
revolved  about  two  axes,  not  in  its  plane,  the  line  will  be 
simultaneously  an  element  of  both  hyperboloids;  that  is,  the 
two  hyperboloids  generated  will  be  tangent.  These,  if  used  as 
the  basis  of  gears,  form  an  ideal  transmission  medium  between 
shafts  not  in  the  same  plane. 


THE  HYPERBOLIC  PARABOLOID 

215.  This  surface  is  generated  by  moving  a  straight  line  along 
two  straight  lines,  not  in  the  same  plane,  keeping  it  always  par- 


\    \    \   \    \   \   \l 


\  \  \  \  \  x  x  \l 


Hyperbolic  Paraboloid,  Showing  Double  Ruling 
FIG.  152. 

allel  to  a  plane  director.     The  elements  are  all  parallel  to  this 
plane,  but  not  to  each  other. 


WARPED  SURFACES  209 

Representation. — If  a  definite  position  of  the  two  directrices  is 
given,  the  elements  may  be  drawn  without  the  assistance  of  the 
plane  director,  by  dividing  each  directrix  into  the  same  number 
of  equal  parts  and  connecting  the  points  in  order.  This  is  so, 
because  a  series  of  parallel  planes  will  cut  any  two  lines  in  pro- 
portional segments. 

216.  Second  Generation. — If  any  two  elements  be  taken,  as 
BD  and  AC  (Fig.  152)  and  divided  as  were  the  directrices  AB 
and  CD,  and  joined  in  order,  the  surface  thus  generated  would  be 
identical  with  the  first  surface.     In  the  two  generations,  the 
directrices  of  the  first  become  elements  of  the  second,  and  the 
elements  of  the  first  become  directrices.     This  makes  it  a  double- 
ruled  surface,  and  the  rule  for  tangents  is  the  same  as  for  the 
hyperboloid. 

217.  It  is  named  the  Hyperbolic  Paraboloid  because  certain 
planes  cut  hyperbolas  from  the  surface,  others  parabolas,  while 
tangent  planes  cut  straight  lines.     It  is  chiefly  of  interest  to  the 
mathematician,  but  it  is  occasionally  met  in  practical  work,  and 
is  then  usually  called  a  "  Warped  Quadrilateral." 

THE  CYLINDROID 

218.  The  Cylindroid  is  a  warped  surface  with  two  curved-line 
directrices  and  a  plane  director.     It  differs  from  a  cylinder  in 
that  its  elements,  while  parallel  to  one  plane  are  not  parallel  to 
each  other.     It  is  found  in  practical  work  in  arches  and  sheet 
metal  work. 

THE  CONOID 

219.  The  Conoid  has  one  curved-line  directrix,  one  straight- 
line  directrix,  and  a  plane  director. 

A  helicoid,  whose  elements  are  at  right  angles  to  the  axis,  may 
be  regarded  as  a  conoid. 

THE  COW'S  HORN 

220.  This  surface  has  three  directrices,  two  being  circles,  and 
•the  third  a  straight  line.     The  circles  are  usually  in  parallel 

planes  with  their  centers  in  a  plane  perpendicular  to  these  par- 
allel planes.  The  straight  line  directrix  usually  lies  in  the  plane 
of  the  centers.  The  Cow's  Horn  is  frequently  encountered  in 
stone  arch  designs,  and  occasionally  in  sheet  metal  hoods, 

14 


210 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


THE  WARPED  CONE 

221.  This  surface,  often  found  in  reducing  hoods  and  pipe  con- 
nections, has  three  directrices;  as  follows,  two  circles  or  ellipses 
(or  one  of  each)  in  planes  not  parallel,  and  the  line  joining  their 
centers,  called  the  axis. 


Elements  Parallel  toH 
Curved  Direc  trix 


Cow's  \\Qrr\(inlsometric) 
Rect.  Directrix 


Axis  is  Rect. 
Directrix 


Warped  Cone 


Curved 
Directrix 


Conoid 


FIG.  153. — Minor  warped  surfaces  and  their  methods  of  generation. 

222.  Other  Warped  Surfaces. — Besides  the  foregoing,  there 
are  many  forms,   which  have  no  specific  names.     They  may 
have    three    straight    directrices,    three    curved    directrices,    or 
combinations  of  these  with  plane  directors.     Whenever  they 
are  met,  they  will  present  no  difficulties  to  the  student,  who  is 
familiar  with  those  described. 

223.  The  Development  of  a   Specimen  Warped   Surface.— 
Develop  the  helicoid  shown  in  Fig.  154,  for  a  half  turn,  between  H 
and  the  helical  directrix. 

1.  Draw  the  elements  Al,  B2,  etc. 

2.  Draw  diagonals  A2,  B3,  etc.,  of  the  quadrilaterals. 

3.  Measure   the  true  lengths   of  the   various   elements   and 
diagonals.     The  construction  on  the  left  shows  a  simple  arrange- 


WARPED  SURFACES 


211 


nient  for  accomplishing  this.  Lay  off  the  heights  of  the  various 
points,  a',  b',  etc.,  on  a  vertical  line,  and  lay  off  the  lengths  of 
the  various  ^/-projections  on  a  horizontal  line,  elements  on  one 


UJ 


side,  diagonals  on  the  other, 
hypothenuses,  a'l  or  a'2,  etc. 

4.  Rectify  the  various  arcs  of  the  helix  and  spiral. 


Their  true  lengths  will  be  the 
The  arcs 


212 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


of  the  helix  are  equal  in  length  to  corresponding  arcs  of  a  circle 

OP 

whose  radius  =  — ^--     (See  the  graphical  solution  given  in  Art. 

COS     v 

204.)  The  lengths  of  the  spiral  arcs  are  obtained  by  stepping  off 
small  chords  with  the  dividers. 

5.  With  these  materials,  construct  the  triangles  in  order,  as 
shown  on  the  right. 

This  method,  with  such  modifications  as  may  be  necessary, 
is  employed  in  the  development  of  all  warped  surfaces.  It  is 
laborious,  but  there  is  no  shorter  method.  The  important 
operation  in  this,  as  in  all  problems  relating  to  surfaces,  is  to 
get  the  elements. 


224. 


EXERCISES 
Helicoid 


Helicoid  dimensions:  pitch  of  helix,  1  in.,  2  in.,  2^  in.,  3  in.;  diameter  of 
helix,  1  in.,  1^  in.,  1^  in.  Angle  of  elements  with  axis,  30°,  45°,  60°,  75°, 
90°. 

834.  Right-hand  helix,  axis  A(2,  0  -  1^)   B(2  +  3  -  1^),  pitch , 

diameter ,  elemental  angle  —  °. 

(a)  Draw  the  projections  of  as  much  of  the  helicoid  as  possible,  showing 

its  fl-trace. 
(6)  Locate  E(1K,  x  —  2)  and  F(2>£  +  1,  y)  on  the  surface. 

835.  Helicoid  in  Ex.  834.     Draw  the  traces  of  the  plane  tangent  at  E  or  F. 

836.  Helicoid  in  Ex.  834.     Locate  the  points  in  which  the  surface  is  pierced 
by  line . 


c: 


Graphic  Layout  No.  64.  —  Piercing  lines  to  use  for  all  the  given  surfaces  in 
these  exercises. 


Dimensioned  Layout  for  the  Same. 

(1)  A(2,  0  -  IK)  B(4  +  2  -  2). 

(2)  C(2,  0  -  2)  B(4  +  2  -  M). 

(3)  E(3  +  2  -  y2)  F(3  +  1  - 

(4)  G(3  +  2  -  2)  J(3  +  K  -  1). 

(5)  M(l  +  1  -  1^)  N(4  +  1  - 


WARPED  SURFACES 


213 


vs 

HtoH 


£i 


HT 
HtoV 


VW 


HW 


Graphic  Layout  No.  65. — Cutting  planes  for  the  surfaces  in  these  exercises. 

Dimensioned  Layout  for  the  Same. 

(1)  S(  +  1)  parallel  to  H. 

(2)  T(  -  1%)  parallel  to  V. 

(3)  P(3  +  3)  3(3  -  3). 

(4)  Q(l  +  3)  1(5  -  2). 

(5)  R(5  •+  1M)  (1  -  3). 

(6)  W(  +  1)  -  (  -  3). 

(7)  X(l  +  3)  5M(4  -  3). 

837.  Helicoid  in  Ex.  834.     Draw  the  projections  of  the  curve  cut  from  the 
surface  by  plane  No. . 

838.  Develop  the    surface  of  Helicoid,  Ex.  834,    through  a  revolution   of 
180°.     (Note. — If  the  elemental  angle  is  more  than  45°,  the  surface 
must  be  limited  by  a  cylinder  of  4-in.  diameter.) 

839.  Left-hand  helix;  axis    M(4,  0  -  \Y±)  N(4  +  3  -  1}£),    pitch , 

diameter ,  elemental  angle °. 

(a)  Draw  the  projections  of  as  much  of  the  helicoid  as  possible,  showing 

//-trace. 
(6)  Locate  E(4^  +  H,  x)  and  K(3,  y  -  2)  on  the  surface. 

840.  Helicoid  in  Ex.  839.     Draw  the  traces  of  the  plane  tangent  at  — 
(E  or  K). 

841.  Helicoid  in  Ex.  839. 
by  the  line . 

842.  Helicoid  in  Ex.  839. 
surface  by  the  plane 

843.  Helicoid  in  Ex.  839. 
(Note. 


Locate  the  points  in  which  the  surface  is  pierced 
Draw  the  projections  of  the  curve  cut  from  the 


Develop  its  surface  through  a  180°  generation. 
If  the  elemental  angle  is  more  than  45°,  the  surface  should  be 
limited  by  a  cylinder  of  4-in.  diameter.) 


Hyperboloid  of  Revolution  of  One  Nappe 

Data  for  Hyperboloid:     Axis,  A  (3,  0  -  1)  B(3  +  3  - 
Circle  Diameter:  ^  in.,  %  in.,  1  in.,  \y±  in.,  1^ 
to  H:  30°,  37^°,  45°,  52^°,  60°. 


Gorge 
Inclination  of  Elements 


214 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


844.  Hyperboloid ;  gorge  diameter ,  elemental  inclination 


°.     Draw 


the  projections  of  its  surface  (24  elements),  and  locate  C(2*^  +  %,  x) 
and  D(3^£,  y  —  2)  on  its  surface. 

845.  Hyperboloid  in  Ex.  844.     Draw  the  traces  of  the  plane  tangent  at 
(C  or  D). 

846.  Hyperboloid  in  Ex.  844.     Locate  the  points  in  which  the  surface  is 
pierced  by  the  line . 

847.  Hyperboloid  in  Ex.  844.     Draw  the  projections  of  the  curve  cut  by 
plane . 

848.  Hyperboloid  in  Ex.  844.     Develop  the  surface  of  the  truncated  portion 
between  the  lower  base  and  the  section  made  by  plane . 

Hyperbolic  Paraboloid. 


* 


«          " 


/N 

2/      I 


>P 

<&*£> 


Graphic  Layout  No.  66. — Directrices  for  paraboloids. 

Dimensioned  Layout  for  Same. 

(1)  A(l  +  2  -  2)  B(4,  0-1)  and  C(3,  0-3)  D(4^  +2-1). 

(2)  E(4  +  1  -  1)  F(4  +  1  -  3)  and  G(l  -  1  -  1)  K(3  -  3  -  3). 

(3)  M(l  +  2  -  1)  N(3  +  2  -  1)  and  O(2  +  3  -  3)  P(3  +  1  -  2). 

(4)  OUK,  0-2)  P(iy2  +  IY2,  0)  and  X(3K  +  2,  0)  Y(3>^,  0  -  1). 

(5)  A(iy2,  0-2)  B(1K  +  2,  0)  and  C(4,  0,  0)  D(4  +  \Y2  -  2). 

(6)  E(l^  +  2  -  Y2)  F(4  +  2  -  2)  and  G(4  +  2,  0)  K(2,  0  -  1). 

(7)  M(1K  +3-1)  N(1K,  0-1)  and  O(4  +  2,  0)  P(4,  0  -  2H). 

(8)  X(l  +  2M  -  H)  Y(3  +  1  -  2)  and  A(3K  +  H  ~  H)   B(5  + 
2  - 


I 

Z 

3 

4|            45° 

5 

V 

/        *? 

10!   e=6o° 

>\             75° 

P 

45° 

\^         £ 

H 

vS'N,' 

1             75° 

Graphic  Layout  No.  67. — Plane  directors. 


WARPED  SURFACES  215 

849.  Hyperbolic  Paraboloid. — Directrices  No.  . 

(a)  Draw  the  projections  of  the  surface  within  the  limits  of  the  given 

directrices,  without  using  a  plane  director. 

(6)  Draw  the  traces  of  its  plane  director. 

(c)  Locate  the  projections  of  the  points  C(2,  x  —  1  K)  and  D(3  +  1,  y) 

on  the  surface. 

850.  Hyperbolic  Paraboloid  in  Ex.  849.     Draw  the  traces  of  a  plane  tan- 
gent to  the  surface  at (C  or  D). 

851.  Hyperbolic   Paraboloid  in  Ex.  849.     Locate  the  point  or  points  in 
which  it  is  pierced  by  the  line . 

852.  Hyperbolic  Paraboloid  in  Ex.  849.     Draw  the  projections  of  the  curve 
cut  from  the  surface  by  plane . 

853.  Hyperbolic  Paraboloid  in  Ex.   849.     Develop  the    surface    included 
between  the  given  directrices. 

854.  Hyperbolic   Paraboloid   in  Ex.   852.     Develop  the  surface  included 
between  the  directrix  —  —  and  the  curve  cut  by  the  given  plane. 

855.  Hyperbolic  Paraboloid. — Directrices  No ,  Plane  Director . 

(a)  Draw  eight  elements  of  the  surface. 

(6)  Locate  the  projections  of  the  points  E(2  -f-  1,  x)  and  F(3,  y  — 
IK)  on  the  surface. 

856.  Hyperbolic  Paraboloid  in  Ex.  855.     Draw  the  traces  of  a  plane  tangent 
to  it  at (E  or  F). 

857.  Hyperbolic  Paraboloid  in  Ex.  855.     Find  the  point  or  points  in  which 
it  is  pierced  by  line . 

858.  Hyperbolic  Paraboloid  in  Ex.  855.     Draw  the  projections  of  the  curve 
cut  by  the  plane . 

859.  Hyperbolic   Paraboloid  in   Ex.   855.     Develop  the  surface  included 
between  the  directrices. 

860.  Hyperbolic   Paraboloid   in  Ex.   855.     Develop  the   surface  included 
between  directrix and  the  cutting  plane . 


MINOR  WARPED  SURFACES 

Conoids,  Directrices,  one  a  Straight  Line,  as  given,  and  either  a  Circle  in 
H,  of  1  in.,  IK  in.,  2  in.,  or  2K  in.  diam.,  or  an  Ellipse  in  H,  1  in.  X  IK  in., 
1  in.  X  2  in.,  or  IK  in.  X  2  in.  Plane  directors  for  (2),  (3),  or  (4),  to  be 
chosen  from  (3)  in  Graphic  Layout  No.  67. 

Cow's  Horn.  Circular  Directrices,  1  in.,  IK  in.,  IK  in.,  1%  in.,  2  in., 
2K  in.,  3  in.,  4  in.  diams.  (Any  combination  may  be.  used.)  Straight 
Directrix,  CD,  lies  in  H. 

Warped  Cone.  Diams.  A,  IK  in.,  IK  in.,  1%  in.,  2  in.,  2K  in.  Diams. 
B,  K  in.,  H  in.,  1  in.,  IK  in.  6  =  22K°,  30°,  37K°,  45°. 


216 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Plane  Di  rector  H 
A  B 


Plane  Director? 
A 


Plane  DirectorlW 
A 


Cylindroids;   Directrices  are  Equal  Semi-Circles 


Plane  Director? 


*"! 


^Conoids*  Rectilinear  Directrix.  Given 

i"  s"  /f-,/// 
Carved  Directrix.  -.Circle  in  }\^Diameter/^t  //}  2,  C£  or 

Ellipse  in  H,  jllgfo'lfa' 
Plane  Director  for  2,3  or  4, Selected  from  Graphic  Layout  67,  Plane  *3 


Cow's  Horn 
Circular  Directrices 
Diams,  iW/jftzfitf 

Straight  Directr!x£Bin\\ 


Warped  Cone 


Graphic  Layout  No.  68. — Minor  warped  surfaces. 


WARPED  SURFACES  217 

861.  Cylindroid  No. ;  diameters  A  and  B,  —  —  (2  in.,  2^  in.,  3  in., 

4  in.),  6  =  -    -  (30°,  45°,  60°,  75°,  90°). 

(a)  Draw  its  projections  (12  elements). 

(6)  Locate  A (2  +  %,  x)  and  B(3K,  Y  -  IK)  on  the  surface. 

(c)  Draw  the  traces  of  a  plane  tangent  at (A  or  B). 

(d)  Locate  the  points  in  which  it  is  pierced  by  the  line . 

(e)  Draw   the   projections  of  the  line   cut  from  its  surface  by  the 
plane . 

(/)    Develop  the  surface  included  between  the  given  bases. 

862.  Conoid  No. ;  diameter  of  circle  in  H ,  plane  director . 

Same  requirements  as  for  Ex.  861. 

863.  Conoid  No. ;  elliptical  base  in  H ,  plane  director . 

Same  requirements  as  for  Ex.  861. 

864.  Cow's  Horn;  diameter  A ,  diameter  B .     Same  requirements 

as  for  Ex.  861. 

865.  Warped  Cone,  No. ;  diameter  A ,  diameter  B ,  6  = . 

Same  requirements  as  for  Ex.  861. 

PRACTICAL  EXERCISES 

In  all  the  development  exercises  use  two  spaces. 

866.  Draw  the  projections  of  a  helical  spring  made  of  24-in.  square  steel, 
outside  diameter  3  in.,  pitch  \}/±  in.,  2  coils. 

867.  Draw  the  projections  of  a  left-handed  square  thread,  2  in.  diameter, 
%-in.  pitch  (single  thread),  and  3  in.  long. 

868.  Draw  the  projections  of  a  helical  spring  made  of  %-in.  wire,  outside 
diameter  2%  in.,  pitch  1  in.,  and  3  coils. 

869.  Draw  the  projections  of  a  V-thread,  2^  in.  diameter,  14-in.   pitch, 
3  in.  long.     Show  a  section  made  by  a  plane  perpendicular  to  the  axis. 

870.  Draw  the  projections  of  a  twist  drill,  1  in.  diameter,  5J^  in.  long,  made 
from  a  flat  bar  ^  in.  thick.     Pitch  of  helix  3^  in.     Sharpen  point  at 
30°.     See  Fig.  155.     Do  not  use  a  ground  line. 

871.  Design  a  screw  conveyor  to  scale,  12-in.  diameter,  capable  of  delivering 
6000  cu.  ft.  of  grain  per  hour,  making  100  revolutions  per  minute. 
Make  the  diameter  of  the  shaft  2  in.     See  Fig.  155. 

872.  Draw  a  conical-helical  spring,  3<£-in.  round  wire,  large  diameter  2^  in., 
small  diameter  \Y±  in.,  3  turns,  1-in.  pitch. 

873.  Draw  the  projections  of  the  pitch  surfaces  of  two  hyperboloidal  gears, 
operating  at  45°,  having  a  gorge  circle  ratio  of  5  to  3. 

Note. — Draw  one  axis  perpendicular  to  H,  and  one  inclined  45°  to  H. 
Take  a  line  for  the  common  element,  at  relative  distances  of  5  and  3  from 
the  axes,  intersecting  their  common  perpendicular. 

874.  Show  the  surface  of  a  cow's  horn  arch,  which  has  one  opening  a  12-ft. 
semicircle  at  A (2,  0,  0)  and  the  other  an  8-ft.  semicircle  at  B(3,  0  —  1). 
Draw  the  projections  of  a  2-ft.  square  stone  on  the  interior  of  the  arch, 
its  center  being  at  C(2%  +  M>  x).     Scale  K  in.  =  1  ft. 

875.  Design  a  conoidal  top  piece  for  the  hip  roof,  as  shown  in  Fig.  155. 
Make  the  three  roofs  45°  to  H,  and  make  the  top  piece  a  right  conoid 
of  suitable  proportions. 


218 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


876.  Develop  the  surface  of  the  automobile  fender  shown  in  Fig.  155. 

877.  Develop  the  surface  of  the  reducing  hood  (warped  cone),  Fig.  155. 

878.  Develop  the  surface  of  Edison  horn  for  talking  machine  (cow's  horn), 
Fig.  155. 


Twist  Drill. 


Conoidal  Top  Piece. 


Screw  Conveyor. 


Reducing  Hood. 
(Warped  Cone) 


-- 510-- " 


Lip  2"wide 
Mud  Guard.N     (Cylindro\d) 


Conical 

Helical  Spring      Horn  forTalking  Machine 
(Con's  Horn) 

FIG.  155. 


879.  What  is  the  surface  of  concrete  form  B,  Fig.  95,  Art.  121?  Develop 
the  surface. 

880.  What  is  the  surface  of  concrete  form  A,  Fig.  95,  Art.  121?     Develop 
the  surface. 


CHAPTER  VIII 

DOUBLE-CURVED  SURFACES  AND  SURFACES  OF 
REVOLUTION 

DOUBLE-CURVED  SURFACES 

225.  A  double-curved  surface  is  generated  by  a  curve  following 
a  curve  in  such  a  way  as  to  generate  a  surface  that  is  neither 
single-curved   or   warped.     A    double-curved    surface    contains 
no  straight  lines.     It  may  be  either  double  convex,  like  a  sphere 
or  ellipsoid,  or  it  may  be  concavo-convex,  like  a  torus,  or  like 
the  turned  surfaces  of  numberless  articles. 

As  all  important  double-curved  surfaces  are  also  surfaces  of 
revolution,  they  will  be  considered  under  that  head. 

SURFACES  OF  REVOLUTION 

226.  Any  line  revolving  about'  an  axis  generates  a  surface  of 
revolution. 

Single-Curved  Surfaces  of  Revolution. — If  the  generatrix  is 
a  straight  line  and  in  the  plane  of  the  axis,  the  surface  will  be 
single-curved.  The  Cylinder  of  Revolution  and  the  Cone  of 
Revolution  are  the  only  examples. 

Warped  Surfaces  of  Revolution. — The  only  example  in  this 
class  is  the  Hyperboloid  of  Revolution  of  One  Nappe. 

Double-Curved  Surfaces  of  Revolution. — Any  plane  curve 
revolving  about  an  axis  generates  this  surface,  except  in  the  case 
of  the  Hyperbola. 

Sphere. — Generated  by  a  circle  revolving  about  one  of  its 
diameters. 

Ellipsoids.  1.  Prolate  Spheroid. — Generated  by  revolving  an 
ellipse  about  its  major  axis.  Its  shape  resembles  a  foot  ball. 

2.  Oblate  Spheroid. — Generated  by  revolving  an  ellipse  about 
its  minor  axis.  Its  shape  resembles  a  door  knob. 

Paraboloids. — Generated  by  revolving  a  parabola  about  either 
axis.  When  revolved  about  the  transverse  axis,  it  generates  the 
surface  used  in  search-light  reflectors.  Why? 

219 


220 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Torus. — Generated  by  revolving  a  circle  about  an  axis  that  is 
not  a  diameter,  usually  completely  outside  the  circle.  In  this 
case  it  is  called  a  ring,  or  annular  torus.  As  a  matter  of  fact,  a 
torus  is  often  generated  by  other  curves  than  the  circle,  or  by  a 
combination  of  curves.  The  architectural  detail  is  of  this 
character.  In  this  work  the  word  torus  will  be  used  to  mean 
the  annular  torus. 

Hyperboloid  of  Revolution  of  Two  Nappes. — Generated  by  re- 
volving a  hyperbola  about  its  transverse  axis.  It  is  not  often 
encountered  in  drafting. 

Many  surfaces  generated  by  combination  curves  will  be  found, 

some  with  names  and  some  without. 
227.  Meridian  and  Right  Sections. 
—Every  plane  perpendicular  to  the 
axis  cuts  a  surface  of  revolution  in  a 
circle.  Hence,  we  may  even  conceive 
a  surface  to  be  generated  by  a  circle 
moving  in  the  direction  of  the  axis, 
and  changing  its  diameter  according 
to  some  law.  Any  surface  turned  in 
a  lathe  is  generated  in  this  way.  (See 
Fig.  156.) 

Every  plane  through  the  axis,  called 
a  Meridian  Plane,  cuts  the  surface 
in  a  Meridian  Line,  which  has  the 
identical  form  of  the  line  used  to 
generate  the  surface. 

Any  two  surfaces  of  revolution  hav- 
ing a  common  axis  will  intersect  in  a 
circle  or  circles,  perpendicular  to  the 

axis.     When  the  axis  is  parallel  to  the  plane  of  projection  these 
intersections  appear  as  straight  lines.     (See  Fig.  156.) 

228.  Representation. — Two  views  of  double-curved  surfaces  of 
revolution  are  all  that  are  necessary,  and  usually  one  is  sufficient, 
and  that  in  outline.  A  center  line  on  V,  perpendicular  to  GL,  is 
almost  always  drawn,  and  the  meridian  section  parallel  to  V 
constitutes  the  F-projection.  The  H -projection  is  merely  the 
circle  of  a  diameter  equal  to  the  largest  dimension  of  the  gener- 
atrix, with  as  many  other  circles,  visible  or  invisible,  as  may  be 
necessary  to  show  the  various  intersections  of  the  surface.  Fig. 
157  shows  a  torus,  which  will  fairly  represent  all  such  surfaces. 


Fio.  156. 


SURFACES  OF  REVOLUTION 


221 


Any  view  of  a  sphere  is  a  circle,  so  both  projections  of  any  sphere 
will  be  circles  of  equal  diameter. 

229.  Problem   81. — To   locate    a    point    on    any   Surface   of 
Revolution. 

Let  it  be  required  to  locate  O  on  the  torus  in  Fig.  157,  assuming 
either  projection. 

Analysis. — Any  plane  parallel  to  //  will  cut  the  torus  in  a 
circle. 

Construction,  assuming  the  F-projection,  ov. 

1.  Through  ov  pass  a  plane 
parallel  to  H.     It  will  cut  a 
circle  of  a  radius  equal  to  OP 
(o>v). 

2.  Draw   the  //-projection 
of  that  circle,  with  its  center 
at  pH. 

3.  Project  ov  to  OH  on  the 
circle. 

Note. — As  ov  could  project  to 
two  different  points  on  the  circle, 
it  is  necessary  to  specify  "front 
surface"  or  "rear  surface"  to 
make  it  exact.  If  O  were  nearer 
the  center  there  would  be  four 
possible  locations,  two  on  the  in- 
side surface.  FIG.  157. 

Construction,  assuming  the  //-projection,  OH. 

1.  Through  OH  draw  an  arc  of  a  circle,  its  radius  equal  to  oHpa. 

2.  Project  01    (where  the   arc   crosses  the  meridian    that  is 
parallel  to  V)  to  the  given  meridian  curve,  at  oV 

3.  Draw  the  trace  of  the  horizontal  plane  through  ovi. 

4.  Project  OH  to  ov  on  this  trace. 

Note. — Two  points  are  possible  in  this  case,  so  "upper  surface"  and 
"lower  surface"  must  be  specified. 

230.  Tangent  Planes. — As  in  other  surfaces,  a  tangent  plane 
to  a  double-curved  surface  is  the  locus  of  all  tangent  lines  at 
any  given  point,  each  tangent  to  one  of  the  infinite  number  of 
curves  of  the  surface  passing  through  that  point. 

The  tangent  plane  to  any  surface  of  revolution  is  perpendicular 
to  the  meridian  plane  containing  the  point  of  tangency. 


222 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


231.  Problem  82. — To  pass  a  plane  tangent  to  a  sphere  at  a 
point  on  the  surface. 

Analysis. — Pass  a  plane  through  the  point  perpendicular  to 
the  radius. 
Let  the  student  make  the  construction. 

232.  Problem  83. — To  pass  a  plane  tangent  to  any  double- 
curved  surface  of  revolution  at  a  given  point  on  the  surface. 

Let  the  given  point  be  O  on  the  surface  of  the  torus,  Fig.  157. 
Analysis. — 1.  Draw  a  tangent  to  the  right  section  through  the 
point. 


FIG.  158. 

2.  Draw  a  tangent  to  the  meridian  section  at  the  point. 

3.  These  two  tangents  will  determine  the  plane. 
Construction. — 1.  Draw  the  tangent  OM  to  the  circle  through 

0,  and  locate  its  F-piercing  point  M. 

2.  Revolve  O  to  OiOiv,  so  that  the  meridian  plane  will  be 
parallel  to  V. 

3.  Draw  the  tangent  Oivniv  to  the  meridian  curve,  in  revolved 
position,  and  locate  its  //"-piercing  point  N. 

4.  Counter-revolve  N  to  its   original  position  nvnH,  into  the 
trace  of  the  meridian  plane  through  0. 

5.  The  two  tangents  will  now  determine  the  tangent  plane, 
HT  being  parallel  to  oHmH,  and  VT  being  drawn  through  my. 


SURFACES  OF  REVOLUTION 


223 


233.  Problem  84. — To  pass  a  plane  tangent  to  a  sphere  through 
a  given  straight  line. 

Analysis. — Regard  the  sphere  as  being  enveloped  by  a  cylinder 
parallel  to  the  given  line,  and  pass  the  plane  tangent  to  the 
cylinder.  The  plane  will  be  tangent  also  to  the  sphere. 

Construction. — Let  AB  (Fig.  158)  be  the  given  line,  and  O  be 
the  center  of  the  sphere. 

1*.  Pass  a  plane,  Q,  through  O  perpendicular  to  AB. 


2.  Find  the  point,  P,  in  which  AB  pierces  Q. 

3.  Revolve  P  and  O  about  the  H-tra.ce  of  Q  into  H  at  pi  and 
QI,  and  draw  a  circle  at  01  of  the  same  radius  as  that  of  the 
sphere. 

4.  Draw  a  tangent  from  pi  to  this  circle. 

5.  Find  its  //-piercing  point  mH  (where  it  crosses  HQ) . 

6.  Draw  the  traces  of  T,  the  required  tangent  plane,  through 
mH  and  the  piercing  points  of  AB. 

v«. 

Note  1. — This  problem  usually  has  two  solutions,  but  if  the  line  AB 
pierces  the  sphere,  the  problem  is  impossible. 

Note  2. — The  projections  of  the  cylinder  were  not  drawn,  as  it  is  merely 
an  imaginary  assistance  to  the  analysis.  It  is  not  even  necessary  to  imagine 


224 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


it,  as  the  normal  plane,  Q,  will  have  to  pass  through  the  center  of  the  sphere, 
although  it  does  not  have  to  be  perpendicular  to  AB.  This  is  convenient, 
however,  and  would  be  necessary  if  the  cylinder  idea  were  carried  out. 

234.  Problem  85. — To  obtain  the  intersection  of  a  plane  with 
any  surface  of  revolution. 

Analysis. — The  simplest  curve  that  can  be  cut  from  a  surface 
of  revolution  is  the  circle,  which  is  cut  by  a  plane  perpendicular 
to  the  axis.  Therefore  a  series  of  planes  perpendicular  to  the 
axis  will  cut  straight  lines  from  the  plane  and  circles  from  the 


Meridian  Method. 


Zone  Method. 
FIG.  160. 

given  surface.  The  curve  of  intersection  will  pass  through  the 
various  intersections  of  the  straight  lines  and  circles.  Fig.  159 
shows  both  projections  of  the  line  of  intersection  of  the  plane 
T  with  a  torus. 

The  vertices  (high  and  low  points)  of  the  curve  may  be  exactly 
located  by  projecting  the  torus  and  plane  on  a  plane  perpendicu- 
lar to  the  .//-trace  of  the  given  plane.  The  vertices  always  lie 
in  the  meridian  plane  which  is  perpendicular  to  the  given  plane. 


SURFACES  OF  REVOLUTION 


225 


Note  1. — Any  plane  will  cut  a  circle  from  a  sphere,  because  any  diameter 
of  the  sphere  may  be  considered  its  axis. 

Note  2. — The  foregoing  method  is  often  useful  in  obtaining  the  in- 
tersections of  planes  with  ruled  surfaces  of  revolution,  particularly  the 
hyperboloid. 

235.  Development  of  Double-Curved  Surfaces  of  Revolution. 

— Development  in  the  strict  sense  is  impossible,  because  there 
are  no  straight  lines.  However,  in  ornamental  sheet  metal  work 
and  the  manufacture  of  globes,  approximate  surfaces  are  made 
from  flat  sheets. 

This  is  done  by  two  methods,  the  "Meridian  Method,"  or 
"Gore  Method,"  in  which  strips  are  cut  along  the  meridian  lines, 
and  the  "Zone  Method"  in  which  belts  of  the  surface,  zones,  are 
regarded  as  conical  surfaces.  Both  methods  are  shown  in  Fig. 
160  and  should  require  no  explanation. 


236. 


EXERCISES 


Sphere        — - 


Prolate  Spheroid 


Ovoid 


CO 


Oblate  Spheroid 


k--A--* 
Torus 


3'lnch  Sphere 
Center  al >M (3,0,0) 

Sphere 


Dim  ens 


_  //     /ft     /  "     7 *'    // 

I  A = c.    £~  P?  2*3 

I     D-/*//*/'"/'4' 
^O/75  ."<     D-/,     *f**,$»'t 


Graphic  Layout  No.  69. — Surfaces  of  revolution.     Axis  O(3,  0  —  1M) 

P(3  +3  —  1H)>  all  surfaces  drawn  %  in.  from  H. 


lcH 


Jd1 


Graphic  Layout  No.  70. — Piercing  lines. 

15 


226 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


Dimensioned  Layout  for  Piercing  Lines. 

(1)  A(2,  0  -  IK)  B(4  +  2  -  2). 

(2)  C(2,  0  -  2)  D(4  +  2  -  Y2). 

(3)  E(3K  +  2  -  Y2)  F(3K  +  1 

(4)  G(3K  +2-2)  K(3K  +  M  -  D. 

(5)  M(l  +  1  -  \Y2)  N(4  +  1  - 

(6)  E(l  +  1  -  K)  F(4  +  1 


vs 

HtoH 


^ 


HT 
BtoV 


VW 


C£i 


HW 


Graphic  Layout  No.  71. — Cutting  planes. 


Dimensioned  Layout  for  Cutting  Planes. 

(1)  S(l  +  1)  (5  +  2)  (1  -  IK)  (5  -  3). 

(2)  T(-lM)||to7. 

(3)  P(3^  +3)3^(3^  -3). 

(4)  Q(l  +3)1(5  -3). 

(5)  R(5  +2)  1(1  -3). 

(6)  W(  +  !)»(-  3). 

(7)  X(  +  lK)»(+3). 

(8)  Y(2+3)  6(5  -  3). 

881.  Sphere  -  ;  A  -  .     Draw  its  projections,  and  -locate 
x)  on  -  (front  or  rear)  surface,  and  D(3;Hj,  y  —  1)  on 
or  lower)  surface. 

882.  Sphere  in  Ex.  881. 

883.  Sphere  in  Ex.  881. 
made  by  plane 


+1> 
(upper 


Find  the  points  in  which  it  is  pierced  by  line  -  . 
Draw  the  projections  and  true  size  of  the  section 


884.  Sphere  in  Ex.  881.     Draw  the  traces  of  the  plane  tangent  at  - 
(C  or  D). 

885.  Sphere  in  Ex.  881.     Draw  the  traces  of  a  tangent  plane  which  contains 
the  line  X  (3  +  3  -  1^)  Y(5  +  1  -  2). 

886.  Sphere  in  Ex.  881.     Draw  the  traces  of  a  tangent  plane  which  contains 
the  line  M(4  +  2  -  1)  N(5  +  1  -  2). 

887.  Sphere  in  Ex.  881.     Develop  the  surface  by  the  Zone  Method. 

888.  Sphere  in  Ex.  881.     Develop  the  surface  by  the  Meridian  Method. 


SURFACES  OF  REVOLUTION  227 

889.  Prolate  Spheroid;  A  • ,  B .     Draw  its  projections,  and  locate 

C(2%,  x  —  1)  on  the (upper  or  lower)  surface,  and  D(3J£  +  1,  y) 

on  the (front  or  rear)  surface. 

890.  Spheroid  in  Ex.  888.     Find  the  points  in  which  it  is  pierced  by  line 


891.  Spheroid  in  Ex.  888.     Draw  the  projection  and  true  size  of  the  curve 
cut  by  plane . 

892.  Spheroid  in  Ex.  888.     Draw  the  traces  of  the  tangent  plane  at 

(C  or  D). 

893.  Spheroid  in  Ex.  888.     Develop  the  surface  by  the  Zone  Method. 

894.  Spheroid  in  Ex.  888.     Develop  the  surface  by  the  Meridian  Method. 

895.  Spheroid  in  Ex.  888.     Draw  the  traces  of  a  tangent  plane  which  con- 
tains M(3  +3-  iyz)  O(4+  1  -  2). 

896.  Spheroid  in  Ex.  888.     Draw  the  traces  of  the  tangent  plane  which  con- 
tains X(4  +  2  -  1)  Y(5  +  1  -  2). 

897.  Oblate  Spheroid ;  A ,  B .     Draw  its  projections,  and  locate 

C(2,  x  —  1)  on  the (upper  or  lower)  surface,  and  D(3%  +  %,  y) 

on  the (front  or  rear)  surface. 

899.  Spheroid  in  Ex.  897.     Draw  the  traces  of  the  plane  tangent  at  — 
(C  or  D). 

900.  Spheroid  in  Ex.  897.     Locate  the  points  in  which  it  is  pierced  by  the 
line . 

901.  Spheroid  in  Ex.  897.     Draw  the  projections  and  true  size  of  the  curve 

cut  by  the  plane . 

902.  Spheroid  in  Ex.  897.     Develop  its  surface  by  the  Zone  Method. 

903.  Spheroid  in  Ex.  897.     Develop  its  surface  by  the  Meridian  Method. 

904.  Torus;  A ,  C .     Draw  its  projections,  and  locate  C(2,  x  —  1) 

on  the (upper  or  lower)  and  D(3J£  +  %>  y)  on  the (front  or 

rear)  surface. 

905.  Torus  in  Ex.  904.     Draw  the   traces  of   the   plane   tangent  at  

(C  or  D). 

906.  Torus  in  Ex.  904.     Locate  the  points  in  which  it  is  pierced  by  the 
line . 

907.  Torus  in  Ex.  904.     Draw  the  projections  and  true  size  of  the  curve 
cut  by  the  plane . 

908.  Torus  in  Ex.  904.     Develop  the  fragment  of  the  surface  — —  (below 
or  above)  the  cutting  plane (by  Zone  Method). 

909.  Torus  in  Ex.  904.     Develop  the  fragment  of  the  surface (below 

or  above)  the  cutting  plane (by  Meridian  Method). 

910.  Ovoid ;  B ,  C .     Draw  the  projections  of  the  surface,  and  locate 

C(2  +  1,  x)  on  the (front  or  rear)  and  D(3%,  y  —  1)  on  the 

(upper  or  lower)  surface. 

911.  Ovoid  in  Ex.   910.     Draw  the  traces  of  the  plane  tangent  at 

(C  or  D). 

912.  Ovoid  in  Ex.  910.     Draw  the  traces  of  a  tangent  plane  which  contains 
P(3  +  3  -  IK)  X(5,  0  -  2). 

913.  Ovoid  in  Ex.  911.     Draw  the  traces  of  a  tangent  plane  which  contains 
Y(3  +  2,  0)  Z(5,  0  -  1%). 


V 


228 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


914.  Ovoid  in  Ex.  911.     Locate  the  points  in  which  the  surface  is  pierced 
by  the  line . 

915.  Ovoid  in  Ex.  911.     Draw  the  projections  and  true  size  of  the  curve 
cut  by  the  plane . 

916.  Ovoid  in  Ex.  911.     Develop  the  fragment  of  its  surface  — : —  (above 
or  below)  the  cutting  plane ,  by (Zone  or  Meridian)  Method. 

917.  Find  the  plane  tangent  to  three  spheres,  all  resting  on  H,  whose  re- 
spective centers  are  A (2  +1-1),  B(3  +  H  -  IK),  and  C(3^  + 

H  -H). 

918.  Develop  the  surface  of  the  ornamental  sheet  metal  surface  of  revolu- 
tion, shown  in  Fig.  161.     Any  convenient  scale. 

Note. — One  strip,  or  meridian,  will  be  sufficient. 


k - : 27-- — 


CHAPTER  IX 

INTERSECTIONS  AND  DEVELOPMENTS  OF  ALL 
SURFACES 

237.  Foreword. — In  order  to  determine  the  line  (or  lines)  of 
intersection  between  two  curved  surfaces,  it  is  necessary  to  find 
a  number  of  lines  in  each  surface  that  will  intersect  those  of  the 
other.     Thus  a  number  of  points,  common  to  both  surfaces,  will 
be  obtained. 

To  obtain  these  intersecting  lines  it  is  necessary  to  pass  a  series 
of  auxiliary  surfaces  through  the  given  surfaces.  It  is  evident 
that  the  most  desirable  lines  to  be  cut  from  the  surfaces  are 
straight  lines  and  circles,  wherever  they  are  possible.  Since 
planes  can  cut  straight  lines  or  circles  from  most  surfaces,  it  is 
evident  then  that  planes  are  chiefly  used  for  this  auxiliary  work. 
The  only  other  surface  used  for  this  work,  in  the  writer's  ex- 
perience, is  that  of  the  sphere,  which  will  cut  circles  from  any 
surface  of  revolution,  whose  axis  passes  through  the  center  of  the 
sphere.  It  is  conceivable  that  other  surfaces  (a  cylinder  or  cone, 
for  example)  might  be  employed,  but  it  is  doubtful  if  it  is  ever 
necessary,  or  even  desirable.  Hence,  the  following  rule  will 
serve  for  general  purposes: 

238.  Rule. — To  obtain  the  line  of  intersection  between  two 
curved  surfaces,  pass  a  series  of  planes  through  the  two  surfaces, 
so  that  their  intersections  with  the  given  surfaces  will  be  the 
simplest  possible  lines.    The  line  determined  by  the  intersec- 
tions of  these  lines,  taken  in  proper  order,  will  be  the  required 
line  of  intersection. 

Important. — Since  lines  that  do  not  lie  on  the  same  surface 
cannot  intersect,  it  is  obvious  that  no  lines  should  be  laid  out  on 
either  of  the  intersecting  surfaces  until  the  cutting  planes  are 
drawn.  Many  students  make  the  error  of  drawing  elements  of 
each  surface,  and  expecting  them  to  yield  the  line  of  intersection, 
when  it  is  more  than  probable  that  very  few  of  the  elements  thus 
drawn  will  intersect.  Follow  the  Rule. 

239.  The  Simplest  Lines. — It  has  been  noted  that  the  simplest 
lines  are  the  straight  line  and  the  circle.     A  straight  line  is  cut 

229 


230 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


from  a  cone  by  passing  a  plane  through  the  apex,  and  from  a  cyl- 
inder by  passing  a  plane  parallel  to  the  axis.  In  the  case  of  a 
cone  intersecting  a  cylinder,  then,  a  natural  procedure  would 
be  to  pass  a  series  of  planes  through  the  apex  of  the  cone  and 
parallel  to  the  axis  of  the  cylinder.  This  is  one  example.  The 
accompanying  table  may  be  of  use  in  deciding  what  sort  of  plane 
to  use  in  the  various  contingencies  that  arise. 


Fia.  162.-|-Showing  straight  lines  cut  from  intersecting  cone  and  cylinder 
by  passing  a  plane  through  the  cone  apex  parallel  to  the  cylinder  axis. 


Surface 
Cone 
Cone 
Cylinder 
Cylinder 
Convolute 
Convolute 
Helicoid 
Warped  surface 
Surface  of  revolution 
Surface  of  revolution 


Intersection 
Straight  line 
Circle 

Straight  line 
Circle 

Straight  line 
Involute 
Straight  line 
Straight  line 
Circle 
Meridian  curve 


Kind  of  Plane 
Through  the  apex. 
Parallel  to  circular  base. 
Parallel  to  the  axis. 
Parallel  to  a  circular  base. 
Tangent  to  helical  cylinder. 
Perpendicular  to  the  axis. 
Through  the  axis. 
According  to  its  generation. 
Perpendicular  to  the  axis. 
Through  the  axis. 


It  may  be  noted  that  planes  cutting  straight  lines  from  warped 
surfaces  are  seldom  easy  to  use  because  of  the  variety  of  their 
directions.  There  are  cases  of  surfaces  having  a  plane  director 
where  it  is  convenient  to  get  straight  line  intersections. 

Frequently  it  is  not  possible  to  use  the  planes  in  the  forego- 
ing list  for  both  of  the  intersecting  surfaces,  but  it  is  always 
possible  to  use  them  for  one.  In  such  cases  the  second  surface 
may  be  cut  in  somewhat  complicated  curves.  It  is  such  cases 
as  these  that  make  it  impossible  to  lay  down  rules  for  all  cases. 


The  draftsman  must  use  his  judgment  and  instinct  for  the  right 
auxiliary  surfaces. 

CONCENTRIC  SPHERES  METHOD 

240.  When  two  surfaces  of  revolution  have  intersecting  axes 
the  easiest  method  of  obtaining  their  intersection  is  to  use  con- 
centric spheres.  By  employing  a  sphere,  as  an  auxiliary  surface, 
with  its  center  at  the  intersection  of  the  axes,  circles  are  cut  from 
the  two  surfaces,  because  the  intersections  of  two  surfaces  of 
revolution  having  a  common  axis  will  be  circles.  A  typical  con- 
struction will  be  shown  in  Problem  89. 


FIG.  163. 
PROBLEMS  IN  INTERSECTIONS 

A  few  representative  problems  of  the  many  that  might  be  given 
are  selected  for  illustration. 

241.  Problem  86. — To  find  the  intersection  between  two  cones. 

Even  in  so  specific  a  problem  as  this,  the  position  and  character 
of  the  two  cones  have  much  to  do  with  the  choice  of  auxiliary 
planes  to  be  used.  We  give  two  representative  cases,  and  both 
could  be  successfully  worked  by  using  another  system  of  auxiliary 
planes. 


232 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


First  Case. — When  the  cones  are  oblique  and  their  bases  are 
circles  in  the  same  plane. 

Analysis. — Pass  a  series  of  planes  parallel  to  the  bases,  inter- 
secting the  cones  in  circles.  Study  Fig.  163  for  the  solution  and 
construction.  Let  the  student  also  work  this  by  passing  planes 
through  the  apices  of  the  cones.  The  advantage  of  the  latter 
method  is  that  it  gives  definitely  the  outside  elements  of  the 
cones  that  are  involved  in  the  intersection. 


c, 


FIG.  164. 

242.  Second  Case. — When  the  cones  are  right  circular,  and 
their  bases  are  in  different  planes. 

Analysis. — 1.  Pass  a  series  of  planes  through  the  apices  of  the 
cones,  cutting  elements  from  each. 

2.  Connect  the  intersections  of  the  elements  in  order. 

Construction. — To  find  the  intersection  between  the  two  right 
circular  cones,  whose  apices  are  A  and  B,  and  whose  bases  are 
in  P  and  H  respectively  (Fig.  164). 


INTERSECTIONS  AND  DEVELOPMENTS  OF  ALL  SURFACES     233 

1.  Connect  A  and  B  with  a  line.     This  line  will  be  contained 
in  all  the  planes  passing  through  the  apices,  and  its  piercing 
points  will  lie  in  all  the  respective  traces  of  these  planes. 

2.  Find  the  .//-piercing  point,  C,  of  the  line  AB,  and  the  P- 
piercing  point,  D,  of  the  same  line. 

3.  Through  CH  draw  any  number  of  traces  of  auxiliary  planes, 
and  draw  the  elements  cut  by  them  from  the  cone  B. 

4.  Draw  the  profile  traces  of  the  same  auxiliary  planes,  and 
draw  the  elements  cut  by  them  from  the  cone  A. 

5.  Locate  the  intersections  of  all  the  elements  with  the  ele- 
ments cut  by  them  from  the  cone  A. 

5.  Locate  the  intersections  of  all  the  elements  with  the  ele- 
ments lying  in  their  respective  planes,  obtaining  the  H-,  V-  and 
P-projections  of  each. 

6.  Draw  the  line  of  intersection  through  them  in  order,  and 
determine  the  visible  and  invisible  parts  of  each  projection  of  the 
curve. 

243.  Problem  87. — To  find  the  intersection  of  two  cylinders. 

This  problem,  like  those  preceding,  may  have  more  than  one 
means  of  solution,  according  to  the  relative  situations  and  char- 
acteristics of  the  cylinders. 

In  the  case  of  the  cylinders  in  Fig.  165,  two  methods  are  avail- 
able, as  follows: 

(A)  Pass  a  series  of  planes  parallel  to  both  axes,   cutting 
elements  from  both  surfaces. 

(B)  Pass  a  series  of  planes  parallel  to  both  circular  bases, 
cutting  circles  from  the  surfaces.     Method  B  can  only  be  em- 
ployed when  the  circular  bases  are  in  the  same  or  parallel  planes, 
as  they  are  in  this  figure. 

In  other  cases,  it  may  be  necessary  to  pass  a  series  of  planes 
parallel  to  one  of  the  planes  of  projection,  or  parallel  to  one  axis 
and  to  the  circular  base  of  the  other  cylinder. 

In  the  specimen  solution,  here  given,  planes  are  passed  par- 
allel to  both  axes. 

Construction. — Let  it  be  required  to  find  the  intersection 
between  the  cylinders  in  Fig.  164,  whose  axes  are  AB  and  CD, 
both  having  circular  bases  in  H. 

1.  Draw  HQ,  passed  through  AB  parallel  to  CD. 

2.  This  trace  intersects  the  base  circles  at  mH,  nH,  OH.  and  pH, 
with  F-projections  at  mv,  nv,  ov,  and  pv  in  GL. 


234 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


3.  Through  M,  N,  O,  and  P  draw  elements  parallel  to  the 
respective  axes. 

4.  These  four  elements  lie  in  the  same  plane,  and  will  intersect 
each  other  in  four  points  on  the  curve.    Locate  these  points. 

5.  Draw  a  series  of  traces  parallel  to  HQ,  and  repeat  operations 
(2),  (3),  (4)  with  each. 

6.  Draw  a  smooth  curve  through  all  of  the  points,  taking  them 
in  order. 

Note. — Do  not  divide  either  circle  into  any  number  of  equal  arcs.     Let 
the  planes  determine  the  elements. 


pr 


FIG.  165. 


244.  Problem  88. — To  find  the  intersection  of  a  cone  and 
cylinder. 

The  following  methods  of  passing  planes  may  be  employed : 

1.  Through  the  apex  of  the  cone,  parallel  to  the  axis  of  the 
cylinder. 

2.  Parallel  to  the  bases  in  any  plane.     (This  method  is  very 
convenient  if  both  surfaces  have  circular  bases  in  H,  V,  or  P.) 

3.  Parallel  to  the  elements  of  the  cylinder  and  perpendicular 
to  the  axis  of  the  cone,  if  it  be  a  right  circular  cone,  if  the  condi- 
tions are  right. 


INTERSECTIONS  AND  DEVELOPMENTS.OF  ALL  SURFACES     235 

Concentric  Sphere  Method. — If  both  are  surfaces  of  revolution 
with  axes  intersecting,  the  process  in  Problem  89  is  usually  the 
best. 

Auxiliary  Plane  Method. — There  are  certain  conditions  when 
the  easiest  and  neatest  solution  is  made  by  projecting  both  sur- 


FIG.  166. 


faces  on  a  plane  perpendicular  to  the  axis  of  the  cylinder.     The 
specimen  solution  given  here  is  for  such  a  condition. 
Construction. — See  Fig.  166. 

1.  Project  the  cone  and  cylinder  on  Q,  perpendicular  to  the 
axis  of  the  cylinder. 

2.  The  various  elements,  1,  2,  etc.,  of  the'cone  will  intersect  the 
circle  in  their  piercing  points  with  the  surface  of  the  cylinder. 


236 


PRACTICAL -DESCRIPTIVE  GEOMETRY 


3.  Project  these  points  back  to  the  H-  and  F-projections  of  the 
elements,  and  draw  the  curve  through  them. 

Note. — Draw  elements  of  the  cone  only. 

Hint. — To  avoid  confusion,  number  your  points  and  elements, 
because  the  following  out  of  the  correct  order  of  these  points  is 
often  very  difficult. 

245.  Problem  89. — To  find  the  intersection  of  any  two  surfaces 
of  revolution  whose  axes  intersect. 


-bv 


8 


FIG.  167. 

Analysis. — 1.  At  the  intersection  of  the  axes  as  a  center  draw 
the  projections  of  a  number  of  concentric  spheres. 

2.  These  spheres  will  cut  circles  from  the  two  surfaces. 

3.  The  intersections  of  the  proper  circles  will  yield  points  on 
the  required  intersection. 


INTERSECTIONS  AND  DEVELOPMENTS  OF  ALL  SURFACES     237 

Construction. — Let  the  circular  cylinder,  whose  axis  is  AB 
(Fig.  167),  and  the  ellipsoid,  whose  axis  is  CD,  be  the  two  surfaces 
of  revolution,  with  axes  intersecting  at  O. 

1.  With  ov  as  a  center,  draw  the   F-projections  of   several 
spheres.     The  F-projections  of  the  circles  cut  by  these  spheres 
will  be  straight  lines.     Why? 

2.  The  circles  cut  from  the  ellipsoid  will  project  on  H  as  circles. 
Draw  these  projections. 

<^>' 


(0 


Q 

UtfM 
W 

R 

U/£l 
(3) 

q 

Lii 

(5) 

k-/;'.-> 

(«) 

U.//'--i 

(<0       (r) 


I   ,//! 
kj*l 


L 

1? 

(18) 

-*-—  J««  H 
M 

(.6) 


,  Graphic  Layout  No.  72. — Bases,  or  Generators,  of  Surfaces,  Spheres, 
Cones,  Cylinders,  Cubes,  Prisms,  Pyramids,  Tori,  Prolate  Spheroids,  Oblate 
Spheroids. 

3.  From  the  F-projections  of  the  intersections  of  the  proper 
circles  project  to  the  circles  1,  2,  3,  etc.     To  do  this  it  will  not 
be  necessary  to  draw  the  H-projections  of  the  cylinder  circles. 
Why? 

4.  Draw  the  two  projections  of  the  lines  of  intersection  through 
these  points  in  order. 

Note  1. — The  elements  of  the  cylinder  are  not  necessary  in  this  solution. 

Note  2. — If  the  position  of  the  surfaces  is  not  as  advantageous  as  in  this 

figure,  project  them  on  an  auxiliary  plane  parallel  to  both  axes.     However, 


238 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


nearly  all  such  problems  are  given  in  this  position,  and  the  difficult  positions 
will  hardly  ever  be  encountered. 


246. 


EXERCISES 


Directions. — In  making  up  the  data  for  the  exercises  from  the  following 
layouts: 

1.  State  what  each  surface  is  to  be,  whether  cone,  sphere,  prism,  torus, 
etc. 

2.  Give  one  of  the  combinations  of  axes. 

3.  Give  the  base  or  generator  of  each  surface.     (Graphic  Layout  No.  72.) 

4.  State  whether  the  base  is  oblique  or  perpendicular  to  the  axis,  if 
necessary. 

5.  Give  the  altitude,  if  necessary,  2  in.,  2^  in.,  3  in. 


«      ! 

-i=t 


A 


-lJ 


7 


LL, 


M 

/   rv 


/ 


Graphic  Layout  No.  73. — Axes  for  intersecting  solids.     (May  be  located 
in  I  or  ///.) 

Data  Specimen  Exercise 

A-Prolate  Spheroid,  B-Oblique  Cone. 
No.  3. 

No.  10.  No.  2. 

Perpendicular.    Base  in  H. 
2^2  in. 

919.-'' Draw  the  line  of  intersection  between  solids  A  and  B,  as  specified  in 
the  data. 

920.  Develop  the  surface  of  solid  A,  as  given  in  Ex.  919. 

921.  Develop  the  surface  of  solid  B,  as  given  in  Ex.  919. 


1.  Surfaces  A^—  B^^. 

2.  Axes-261-.  '  Q 

3.  Base  or  Generator, — — ,     ^  . 

4.  Oblique  or  Perpendicular-*- — , 

5.  Altitude— — ,   ?  *•  . 


INTERSECTIONS  AND  DEVELOPMENTS  OF  ALL  SURFACES     239 

PRACTICAL  EXERCISES  IN  VARIOUS  SURFACES 

922.  Draw  the  line  of  intersection  between  the    3-in.    sphere,    center   at 
O(3  +  IK  -2)  and  a  1%-in.  circular  cylinder,  axis  A(l  +  IK  —  2) 


A 

B 

r        t 

D 

E-F 

G 

20 

20 

8 

12 

8 

16 

30 

28 

9 

14 

9 

Id 

40 

36 

10 

16 

12 

21 

50 

40 

12 

20 

14 

28 

Loading  5kip  for  Mixer 


K iO-—-->\     K 7-- 

Coffee  Receiver 


A 

B 

c 

D 

E 

Id 

40 

10 

10 

16 

20 

4O 

10 

10 

16 

24 

SO 

1?. 

12 

20 

30 

40 

15 

16 

26 

Concrete  Hopper 


A 

B 

R 

R' 

9 

5 

4'/2 

B 

/2 

12 

6 

10 

16 

16 

8 

14 

20 

20 

10 

18 

Cylindroidal  Hood 


S 


A 

B 

C 

D 

^0 

3 

14 

£ 

J<? 

50 

20 

5 

40 

54 

30 

9 

48 

72 

40 

JO 

60 

96 

44 

14 

Fan  Transition  Piece 


k <S H 

Gasoline  Measure 


Fia.  168. — Exercises  in  various  surfaces. 


240 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


923.  Lay  out  the  patterns  for  the  Loading  Skip  for  Mixer,   No. ,  Fig. 

168. 


A     ' 

B 

c 

D 

•     E 

F 

6  ' 

20      • 

24 

10  * 

36 

'     e     . 

-    it 

//  ' 

30 

30 

/«  • 

45 

10 

n 

IS 

40 

44 

10 

.    60 

11 

fi 

18 

7? 

80 

'      30 

IOS 

10 

/a 

28 

Concrete  Mixer  Hopper 


_  _.!L__  _     • 


K L A—-:-- >) 

Octagonal  Dome 

A=  /Off,/?  ft,  !6Ft,  ZOFt,  Z4rf,  40Ff,'50ft,60ff. 


'A -- 

•EllipSOiQlal  Domte 


Handlers" 


k-e-'-H 

Connecting  Rod  End. 


3 

3  If 

L<*   <£>  T 

U  ;  V 

... 

Sugar  Scoop. 


K"4."H 

Octagonal    Tie    Rod. 
FIG.  169. — Practical  exercises. 

924.  Lay  out  the  patterns  for  the  Cylindroidal  Hood,  No. ,  Fig.  168. 

925.  Lay  out  the  patterns  for  the  Receiver  for  Coffee  Grinder,  Fig.  168. 


INTERSECTIONS  AND  DEVELOPMENTS  OF  ALL  SURFACES     241 


926.  Lay  out  the  patterns  for  the  Fan  Transition  Piece,  Fig.  168. 

927.  Lay  out  the  patterns  for  the  Concrete  Hopper ,  Fig.  168. 


Two-Way  Boiler  Breeching 

HJ'"!S      •••H'^r 


Oblong  Outlet  for  Boiler 


Smoke  Connector  for  Two  Boilers 


Kewaunee   Flue  Attachment 
'-o'Diam;>\ 


1    1-1  ?"  n 
k -S-3-— 

Breeching  for  Kewaunee  Boilers 


Tudor  Boiler  Breeching 


3-0  < 


Oval  3-Way  Breeching 


Stack  Connection- Ames  Boiler 


FIG.  170. — Boiler  problems. 


928.  Lay  out  the  patterns  for  the  Concrete  Mixer  Hopper ,  Fig.  169. 

Dimensions  given  in  inches  in  the  table. 

929.  Develop  the  surface  of  the  Octagonal  Dome,  Fig.  169.     Let  A  =  — 
and  B  =  —  — .     One  strip  is  sufficient. 


242  PRACTICAL  DESCRIPTIVE  GEOMETRY 

930.  Develop  the  surface  of  the  Ellipsoidal  Dome,  Fig.  169.     Let  A  = , 

and  B  =  . 

931.  Lay  out  the  patterns  of  the  Sugar  Scoop,  Fig.  169. 

932.  Draw  the  intersection  of  surfaces  in  the  Connecting  Rod  End,  Fig.  169. 

933.  Draw  the  line  of  intersection  of  the  surfaces  in  the  Octagonal  Tie  Rod, 
Fig.  169. 

934.  Lay  out  the  patterns  for  the  Two-way  Boiler  Breeching,  Fig.   170. 
Supply  the  necessary  dimensions. 

935.  Lay  out  the  patterns  for  the  Smoke  Connector  for  Two  Boilers,  Fig. 
170. 

936.  Lay  out  the  patterns  for  the  Oblong  Outlet,  Fig.  170. 

937.  Lay  out  the  patterns  for  the  Kewaunee  Flue -Attachment,  Fig.  170. 

938.  Lay  out  the  patterns  for  the  Breeching  for  Kewaunee  Boilers,  Fig.  170. 

939.  Lay  out  the  patterns  for  the  Tudor  Boiler  Breeching,  Fig.  170.     Supply 
the  necessary  dimensions. 

940.  Lay  out  the  patterns  for  the  Oval  Three-way  Breeching,  Fig.   170. 
Supply  the  necessary  dimensions. 

941.  Lay  out  the  patterns  for  the  Ames  Boiler  Stack  Connection,  Fig.  170. 


CHAPTER  X 

PICTORIAL  PROJECTION 
PERSPECTIVE 

247.  One  of  the  most  important  and  interesting  applications 
of  Descriptive  Geometry  is  the  art  of  Perspective,  or,  as  it  is 
sometimes  called,  Conical  Projection.  As  was  pointed  out  in  the 
beginning  of  the  work,  Perspective  Projection  differs  from 
Orthographic  in  that  it  represents  the  object  by  an  image  which 
is  practically  identical  with  that  formed  on  the  retina  of  the  eye 
of  the  observer.  The  title,  Conical  Projection,  is  a  correct  one, 
since  the  picture  is  made  by  the  projection  of  rays  that  converge 
in  a  point. 

Numerous  methods  are  given  by  various  writers  of  text-books, 
and  educators  make  use  of  all  of  them.  The  method  presented 
here  is  offered,  for  two  reasons:  (1)  its  extreme  simplicity  and 
reasonableness;  and  (2)  because  it  is  a  pure  application  of  Descrip- 
tive Geometry,  and,  being  such,  it  can  be  given  to  the  student 
already  grounded  in  the  fundamentals  without  much  detailed 
illustration  and  complicated  directions. 


Observer 


Object 


248 


Picture  Plane 
FIG.  171. — How  the  pictorial  image  is  made. 

CONICAL  PROJECTION 


If  a  plane,  called  the  Picture  Plane,  be  so  placed  as  to  intersect 
the  rays  coming  from  an  object  to  the  eye,  the  piercing  points  of 

243 


244  PRACTICAL  DESCRIPTIVE  GEOMETRY 

the  rays  will  yield  a  perfect  visual  image  of  the  object.  This  is 
the  exact  process  of  the  camera,  so  that  a  photograph  is  an  ap- 
proximately perfect  perspective. 

249.  The  Construction  of  a  Perspective  from  a  Working 
Drawing. 

In  Fig.  172  a  set  of  steps  is  shown  in  third  angle  projection, 
somewhat  below  the  horizon. 

Note. — The  horizon  should  not  be  confused  with  GL,  as  it  has  not  the 
same  function. 

The  point  of  view  for  the  perspective  may  be  taken  from  any 
convenient  position,  but  it  is  customary  to  view  the  object  "from 


Horizon 


Resting ''Surface 
FIG.  172. — Object  in  orthographic  projection. 

an  angle,"  so  as  to  show  two  sides.  This  takes  away  the  uncom- 
promising appearance  of  the  front  elevation,  as  it  appears  in  the 
working  drawings,  and  offers  opportunity  to  the  draftsman  of 
taking  the  most  interesting  view.  In  making  interior  views  of 
buildings,  and  in  stage  scenery,  etc.,  the  point  of  sight  is  often 
taken  directly  in  front  of  the  scene. 

To  draw  the  picture  of  the  steps,  place  the  plan  and  elevation 
(or  side  view,  if  preferred)  in  a  convenient  position  on  the  drawing, 
with  the  horizon  at  its  proper  elevation;  see  Fig.  173.  Locate 
the  point  of  sight,  SH,  sv,  at  a  suitable  position,  where  both  front 
and  side  are  nicely  visible.  Note  that  sv  is  always  on  the  horizon. 
The  picture  plane,  HP,  is  placed  so  as  to  give  the  desired  scale, 
and  should  be  about  perpendicular  to  the  central  line  of  sight. 
The  scale  is  determined  by  the  ratio  of  the  distances  of  the  object 


PICTORIAL  PROJECTION  PERSPECTIVE 


245 


and  the  picture  plane  from  the  eye.  In  this  figure,  the  ratio  is 
about  3  :  4,  and  the  picture  is  therefore  about  three-quarter 
scale.  If  an  enlargement  is  desired,  place  HP  beyond  the  plan 
of  the  object. 

For  convenience,  the  picture  plane  is  taken  as  a  vertical  plane 
(even  though  the  central  line  of  sight  may  not  be  horizontal); 
therefore  HP,  its  H-tra.ce,  contains  the  H -projections  of  all  its 
points.  Thus  the  piercing  points  of  the  rays  are  easily  deter- 
mined, since  they  are  projected  on  the  //-trace. 


Plan 


Elevation  V 

FIG.  173. — Drawing  a  perspective  from  an  orthographic  projection. 

By  connecting  sv  with  all  points  in  the  elevation,  and  SH  with 
all  points  in  the  plan,  the  H-  and  F-projections  of  the  projecting 
cone  are  drawn.  The  section  of  this  cone  made  by  the  picture 
plane  is  the  picture.  Since  the  picture  plane  is  not  parallel  to 
V,  the  picture  is  not  shown  in  true  size,  unless  revolved  parallel 
to  F. 

In  Fig.  173,  the  F-projection  of  the  conic  section  made  by  the 
picture  plane  is  shown  in  dotted  lines.  By  revolving  HP  to  HP' 
(parallel  to  F)  the  picture  is  shown  in  true  size  in  solid  lines. 

250.  To  Construct  a  Perspective  of  an  Object  and  its  Shadow. 

The  process  in  this  case  differs  in  no  way  from  that  of  the  pre- 
ceding problem.  The  intersection  of  the  cone  by  the  picture 
plane  is  made,  and  then  revolved  parallel  to  F. 


246 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


FIG.  174. — Perspective  of  an  object,  showing  its  shade  and  shadow. 


n 


Horizon 


n   n  n    n 


Projections  of  a  Bungalow 
FIG.  175. 


PICTORIAL  PROJECTION  PERSPECTIVE 


247 


As  an  illustration  of  the  application  to  more  complicated  ob- 
jects, a  bungalow  is  shown  in  Fig.  176,  drawn  in  perspective  by 


\L     -y* 


o 

I 

3 

.C 


conical  projection  from  the  working  drawing  as  given  in  Fig.  175. 
In  Fig.  176  only  a  few  construction  lines  are  shown. 


248 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


251. 


EXERCISES 


Draw  the  perspective  of  the  following  objects,  to  the  scale  designated  by 
the  instructor,  one-half,  three-quarters,  full,  or  double  the  scale  of  the  work- 
ing drawings.  Make  the  working  drawings  as  large  as  the  space  will  allow 


1  i 

3 

Core    BOA. 


Graphic  Layout  No.  74. 


Bearing 


to  give  plenty  of  room  for  the  perspective  and  its  construction.     Place 
all  small  objects  below  the  horizon. 
942.  1^-in.  cube.     Scale . 


Graphic  Layout  No.  76. 

943.  Frustum  of  a  square  pyramid,  lower  base  1^  in.,  upper  base  %  in., 
altitude  (base  to  base)  1  in.     Scale . 

944.  Wooden  box  20  in.  long,  12  in.  wide,  6  in.  deep,  walls  1  in.  thick,  with- 
out top.     Scale . 

945.  Place  a  1^-in.  equilateral  triangular  prism,   !}>£  in.  long,  on  a  1-in. 


PICTORIAL  PROJECTION  PERSPECTIVE  249 

cube   (resembling  a  house  with  gable  roof).     Draw  the  shades  and 
shadows,  and  their  perspective.     Scale . 

946.  Place  a  1^-in.  square  pyramid,  \^-\i\.  altitude,  on  a  1-in.  cube.     Draw 
the  shades,  shadows,  and  perspective.     Scale  . 

947.  Core  box  for  octagonal  core,  Graphic  Layout  No.  74.     Scale . 

948.  Bearing  box,  Graphic  Layout  No.  74.     Scale . 

949.  House,  Graphic  Layout  No.  75.     Scale . 

950.  House,  Graphic  Layout  No.  75.     Shades,  shadows,  and  perspective. 
Scale . 

951.  Draw  the  perspective  and  the  shadow,  of  the  object,  or  objects,  in 
Ex. (selected  from  714-744,  Chap.  IV). 


PSEUDO-PERSPECTIVES 

ISOMETRIC,  CAVALIER,  AND  CABINET  PROJECTIONS 

252.  For  certain  purposes  mechanical  variants  on  perspective 
are  preferred  to  conical  projection.  These  variants  are  collect- 
ively called  Single  Plane  Projection.  The  system  of  outstand- 
ing preference  of  these  is  called  Isometric  Projection,  which  means 
' '  E  qual  M  easure ' '  pro]  ection.  It  is  orthographic  proj  ection  on  an 
oblique  plane,  as  will  be  shown.  Isometric  has  its  advantages 
and  objections  compared  with  perspective.  Its  advantages  are: 
(1)  ease  and  rapidity  of  execution,  its  principal  lines  being  laid 
off  with  the  30°  triangle;  (2)  measurements  of  principal  lines  are 
to  scale,  which  facilitates  drawing  and  makes  it  more  valuable  for 
shop  purposes;  (3)  there  are  engineering  advantages  in  isometric 
layouts  of  wiring  and  piping  jobs.  It  has  been  called  "Practical 
Perspective"  by  some  writers.  Its  chief  and  only  serious  ob- 
jection is  that  it  does  not  take  distance  into  account,  and  there- 
fore there  is  a  distortion,  which  gives  an  unpleasant  appearance 
•when  the  object  is  long  in  recession. 

ISOMETRIC  PROJECTION 

Construction. — Place  a  cube  so  that  its  base  diagonal  is  parallel 
to  V;  see  Fig.  177.  Pass  a  T7-projecting  plane  T  through  A  and 
B,  and  project  the  plane  on  R  drawn  parallel  to  T.  Its  R-pro- 
j ection  will  be  isometric;  that  is,  the  several  edges  will  all  be  of 
equal  length,  as  they  are  in  reality,  and  the  projection  will  show 
three  faces  in  one  view.  The  isometric  projections  of  these  edges 

are  •,     '  „„  of  their  true  length.     The  edge?  of  the  cube  in  this 


250 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


projection  run  in  three  directions  at  equal  angles  (120°) .  Thus,  in 
Isometric,  the  90°  angles  of  the  three  systems  of  parallels  of  the 
cube  (called  the  isometric  axes)  radiate  at  120°  from  each  other. 


FIG.  177. 


ISOMETRIC  DRAWING 


263.  Isometric  Drawing  differs  from  isometric  projection  only 
in  that  the  latter  would  require  all  "isometric  lines"  to  be  drawn 

(81  \ 
TT^J  scale.    Since  most  working  drawings  are  them- 


selves drawn  to  a  reduced  scale,  and  the  isometric  is  chiefly 
for  pictorial  purposes,  this  distinction  is  of  no  importance,  and 
the  isometric  lines  may  be  drawn  the  full  length  of  the  working 
drawing. 

ISOMETRIC  AXES 

Lay  off  AO,  BO,  and  CO,  as  shown  in  Fig.  178,  radiating  at 
120°,  AO  and  CO  being  drawn  at  30°  from  the  horizontal.  These 
three  lines  represent  three  of  the  adjacent  edges  of  the  cube,  BO 
a  vertical  and  AO  and  CO  horizontal  lines  at  right  angles  to 
each  other.  In  isometric  all  objects  are  drawn  so  that  the 
rectangular  lines  run  in  these  directions.  Other  lines  (non- 
isometric)  are  laid  off  by  joining  points  on  the  isometric  lines,  or 
by  erecting  ordinates  from  known  points. 


PICTORIAL  PROJECTION  PERSPECTIVE 

254.  Isometric  drawing  of  a  cube. 
Construction. — Given  the  cube  in  Fig.  179. 
1.  Lay  off  the  isometric  axes,  as  in  Fig.  178. 


251 


FIG.  178. — Isometric  axes. 

2.  Make  each  edge,  AO,  BO,  CO,  equal  to  the  true  length  of 
the  cube  edges. 

3.  Complete  the  three  parallelograms  on  these  edges. 


Orthographic  Projection  of  a  Cube. 
FIG.  179. 


Isometric  Drawing 
of  the  same  Cube. 


FIG.  180. 


255.  To  locate  a  point  isometrically. 

Refer  to  Figs.  179  and  180.  Let  M,  inside  the  cube,  be  the 
required  point,  given  by  its  projections  mH,  mv. 

1.  Draw  isometric  ordinates,  as  follows:  (a)  a  perpendicular 
MN  to  the  face  AOBD;  (6)  a  perpendicular  NP  to  the  edge  DB. 


252 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


2.  Lay  off  these  ordinates  in  their  true  length  on  the  proper 
isometric  lines  in  Fig.  180,  thus  locating  M  in  its  proper  place 
within  the  cube. 

255.  To  draw  an  exact  circle. 

Construction. — Refer  to  Figs.  181  and  181a.  On  the  left  face 
of  the  cube  (Fig.  181)  is  inscribed  a  circle,  with  squares  in  each 
corner  touching  the  circle  at  E,  G,  H,  and  K. 

1.  Draw  each  of  these  squares  in  isometric,  Fig.  181a. 

2.  Draw  an  ellipse  in  the  face  of  the  cube,  Fig.  181a,  touching 
E',  G',  Hr,  and  K',  and  the  center  point  of  each  of  the  four  edges. 


T    P 


FIG.  181. 


256.  To  draw  an  approximate  circle. 
Construction. — Refer  to  Fig.  181a. 

1.  Connect  S'  and  T'  with  the  middle  points  of  the  opposite 
edges. 

2.  Using  T'U'  as  a  radius,  draw  arcs  from  T'  and  S'  as  centers, 
from  middle  point  to  middle  point. 

3.  Using  Q'  and  R'  as  centers,  and  Q'U'  as  radius,  complete 
the  oval. 

Note. — This  system  of  drawing  the  circle  isometrically  is  almost  exclu- 
sively employed  in  commercial  drafting,  since  it  can  be  done  with  the  com- 
passes and  is  clean  cut  in  appearance. 

257.  To  draw  lines  making  any  desired  angle  with  any  given 
line. 

Construction. — Required  to  draw  lines  at  30°,  45°,  and  75°  to 
LM,  Fig.  181. 

1.  Draw  LN,  LO,  and  LP  in  Fig.  181  at  their  true  angles  to 
LM. 

2.  From  M'  (Fig.  181a)  lay  off  the  ordinates  M'N'  and  M'O' 
equal  to  MN  and  MO. 


PICTORIAL  PROJECTION  PERSPECTIVE 


253 


3.  From  T '  lay  off  T'P'  equal  to  TP.  The  lines  L'N',  I/O',  L'T' 
will  be  the  required  lines  in  isometric. 

Note. — If  the  angles  were  laid  off  at  M,  instead  of  L,  they  would  work  out 
differently.  The  corners  at  M  and  L  are  each  90°,  but  in  isometric  one  is 
60°  and  the  other  120°,  both  representing  the  same  angle. 

258.  To  draw  the  frustum  of  a  hexagonal  pyramid. 

Construction. — Exactly  as  in  previous  exercises.  Fig.  182  is 
the  working  drawing  and  Fig.  182a  the  isometric.  Construction 
lines  are  shown. 


br— - 


FIG.  182. 


Note. — Do  not  draw  in  the  invisible  lines.  They  are  drawn  here  to  make 
the  figure  complete,  but  in  practical  work,  where  isometric  drawings  are 
used,  the  dotted  lines  are  seldom  drawn. 

259.  Other  Methods  of  Pictorial  Projection. 

Under  certain  conditions,  a  closer  approximation  of  a  per- 
spective than  that  shown  by  the  isometric  is  obtained  by  the  use 
of  other  oblique  projections,  called  Cavalier  and  Cabinet  Pro- 
jections. 

Cavalier  Projection. — The  object  is  placed  with  one  face 
parallel  to  the  picture  plane,  and  the  projectors  are  considered 
inclined  45°  to  the  picture  plane.  Thus  one  face  is  the  same  as 
the  orthographic  view  would  be,  and  all  lines  perpendicular  to 
the  picture  plane  are  projected  in  their  true  length.  It  thus 
becomes  a  form  of  isometric  projection,  with  less  distortion, 


254 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


because  one  face  is  the  same  as  othographic  projection  would 
make  it. 

Cabinet  Projection. — To  further  reduce  the  distortion,  the 
receding  lines  are  shortened  one-half.  Otherwise  the  drawing  is 
made  like  cavalier  projection. 


Cavalier  Projection 
FIG.  183. 

Fig.  183  shows  a  box  drawn  by  the  cavalier  method,  and  Fig. 
184  shows  the  same  box  drawn  by  the  cabinet  and  isometric 
methods. 


Cabinet  Projection 


Isometric  Drawing 


FIG.  184. 


EXERCISES  IN  SINGLE  PLANE  PROJECTION 

260.  Make (isometric,  cavalier,  cabinet)  drawings  of  the  following 

object,  making  them  to  reduced  scale,  if  necessary. 

952.  Wooden  box,  without  cover,  10  in.  long,  6  in.  wide,  4  in.  deep,  made  of 
%-in.  boards.     Stand  on  end,  open  to  the  front. 


255 


i 

j-|i[\_ 

sSSrirrii  ^ssl 
LLixJ 

==?• 

- 

k 

L 

i 

!     ., 

^ 

i 
i 

^ 
pi 

•! 

? 

~ 

\ 

fl 

- 

7 

.-> 

;; 

<- 

?g» 
32 

• 
i 

i 

! 

f          3 

C_ 

j 

Pulley 


Hexagonal 
Bolt. 


Pipe    Tee.. 


_j^_         ->i;  <---„- //g"- >|;k-  '/- 

ix'->l  IK  -  '•--  ^       Corners  /*  ff. 


K- /2- >1 

Mission  Book  Rack.    , 


Crank  Forging. 


Piilow  Block. 


FIG.  185. 


256 


PRACTICAL  DESCRIPTIVE  GEOMETRY 


953.  Same  box,  in  natural  position,  with  a  cover  open  120°. 

954.  Pentagonal  plinth,  2-in.  diameter,  ^  in.  thick,  surmounted  by  a  1-in. 
cylinder,  1%  in.  high. 

955.  Bolt  shown  in  Fig.  185. 

956.  Pulley,  Fig.  185. 

957.  Book  rack,  Fig.  185. 


958. 


-in.  pipe  tee,  Fig.  185. 


I  "Elbow 


I  Elbow 


FIG.  186. 

959.  Bearing  brass,  Fig.  185. 

960.  Crank  forging,  Fig.  185. 

961.  Pillow  block,  Fig.  185. 

962.  Frustum  of  pyramid,  given  in  Ex.  943. 

963.  Saw  horse,  Fig.  86,  Chap.  III. 

964.  Stand,  Fig.  89,  Chap.  III. 

965.  Piping  layout,  Fig.  186.      This  is  intended  for  a  large  sheet,  and  makes 
a  fine  drawing  if  traced. 

Note. — First  make  the  layout  of  the  center  lines,   making  necessary 
allowances  for  all  connections. 


FIG.  187. — A  problem  in  sheet  metal  work,  involving  nearly  every  prob- 
lem in  Descriptive  Geometry,  including  the  intersection  and  development 
of  plane,  single-curved,  and  warped  surfaces.  The  picture  itself  is  a  problem 
in  shades,  shadows,  and  perspective. 


i 

)  7  /  - 

Date  Due 


' 


NOV  25|l960 

FEB     31961 

NOV  3      J96i 

NOV10  1 

961 

c,ni  1  2  -62 

ninu  ft      Ifl 

n 

NUV  o     *9w 

OCT  161J 

64 

DEC  & 

1966^ 

JAN  2 

5  1967 

DEC  13 

1967> 

i- 

NPV 

(  1  1971 

^16  'ft 

due  eirtlt 
fiMtftM 

! 

®) 

PRINTEC/ 

JN  U.  S.  A. 

The  RALPH  D.  REED  LIBRARY 

DKPARTMRNT  WK  «BOU»GY 

UNIVERSITY  of  CAt.lFORNIA 

UW  ANGELB8.  CAMP. 


000  584 


